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krahets
@@ -190,25 +190,180 @@ comments: true
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=== "Go"
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```go title="n_queens.go"
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[class]{}-[func]{backtrack}
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/* 回溯算法：N 皇后 */
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func backtrack(row, n int, state *[][]string, res *[][][]string, cols, diags1, diags2 *[]bool) {
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// 当放置完所有行时，记录解
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if row == n {
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newState := make([][]string, len(*state))
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for i, _ := range newState {
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newState[i] = make([]string, len((*state)[0]))
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copy(newState[i], (*state)[i])
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[class]{}-[func]{nQueens}
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}
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*res = append(*res, newState)
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}
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// 遍历所有列
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for col := 0; col < n; col++ {
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// 计算该格子对应的主对角线和副对角线
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diag1 := row - col + n - 1
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diag2 := row + col
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// 剪枝：不允许该格子所在 (列 或 主对角线 或 副对角线) 包含皇后
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if !((*cols)[col] || (*diags1)[diag1] || (*diags2)[diag2]) {
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// 尝试：将皇后放置在该格子
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(*state)[row][col] = "Q"
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(*cols)[col], (*diags1)[diag1], (*diags2)[diag2] = true, true, true
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// 放置下一行
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backtrack(row+1, n, state, res, cols, diags1, diags2)
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// 回退：将该格子恢复为空位
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(*state)[row][col] = "#"
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(*cols)[col], (*diags1)[diag1], (*diags2)[diag2] = false, false, false
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}
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}
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}
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/* 回溯算法：N 皇后 */
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func backtrack(row, n int, state *[][]string, res *[][][]string, cols, diags1, diags2 *[]bool) {
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// 当放置完所有行时，记录解
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if row == n {
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newState := make([][]string, len(*state))
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for i, _ := range newState {
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newState[i] = make([]string, len((*state)[0]))
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copy(newState[i], (*state)[i])
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}
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*res = append(*res, newState)
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}
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// 遍历所有列
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for col := 0; col < n; col++ {
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// 计算该格子对应的主对角线和副对角线
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diag1 := row - col + n - 1
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diag2 := row + col
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// 剪枝：不允许该格子所在 (列 或 主对角线 或 副对角线) 包含皇后
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if !((*cols)[col] || (*diags1)[diag1] || (*diags2)[diag2]) {
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// 尝试：将皇后放置在该格子
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(*state)[row][col] = "Q"
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(*cols)[col], (*diags1)[diag1], (*diags2)[diag2] = true, true, true
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// 放置下一行
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backtrack(row+1, n, state, res, cols, diags1, diags2)
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// 回退：将该格子恢复为空位
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(*state)[row][col] = "#"
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(*cols)[col], (*diags1)[diag1], (*diags2)[diag2] = false, false, false
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}
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}
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}
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func nQueens(n int) [][][]string {
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// 初始化 n*n 大小的棋盘，其中 'Q' 代表皇后，'#' 代表空位
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state := make([][]string, n)
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for i := 0; i < n; i++ {
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row := make([]string, n)
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for i := 0; i < n; i++ {
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row[i] = "#"
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}
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state[i] = row
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}
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// 记录列是否有皇后
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cols := make([]bool, n)
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diags1 := make([]bool, 2*n-1)
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diags2 := make([]bool, 2*n-1)
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res := make([][][]string, 0)
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backtrack(0, n, &state, &res, &cols, &diags1, &diags2)
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return res
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}
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```
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=== "JavaScript"
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```javascript title="n_queens.js"
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[class]{}-[func]{backtrack}
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/* 回溯算法：N 皇后 */
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function backtrack(row, n, state, res, cols, diags1, diags2) {
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// 当放置完所有行时，记录解
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if (row === n) {
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res.push(state.map((row) => row.slice()));
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return;
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}
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// 遍历所有列
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for (let col = 0; col < n; col++) {
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// 计算该格子对应的主对角线和副对角线
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const diag1 = row - col + n - 1;
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const diag2 = row + col;
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// 剪枝：不允许该格子所在 (列 或 主对角线 或 副对角线) 包含皇后
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if (!(cols[col] || diags1[diag1] || diags2[diag2])) {
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// 尝试：将皇后放置在该格子
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state[row][col] = 'Q';
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cols[col] = diags1[diag1] = diags2[diag2] = true;
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// 放置下一行
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backtrack(row + 1, n, state, res, cols, diags1, diags2);
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// 回退：将该格子恢复为空位
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state[row][col] = '#';
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cols[col] = diags1[diag1] = diags2[diag2] = false;
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}
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}
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}
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[class]{}-[func]{nQueens}
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/* 求解 N 皇后 */
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function nQueens(n) {
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// 初始化 n*n 大小的棋盘，其中 'Q' 代表皇后，'#' 代表空位
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const state = Array.from({ length: n }, () => Array(n).fill('#'));
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const cols = Array(n).fill(false); // 记录列是否有皇后
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const diags1 = Array(2 * n - 1).fill(false); // 记录主对角线是否有皇后
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const diags2 = Array(2 * n - 1).fill(false); // 记录副对角线是否有皇后
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const res = [];
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backtrack(0, n, state, res, cols, diags1, diags2);
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return res;
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}
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```
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=== "TypeScript"
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```typescript title="n_queens.ts"
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[class]{}-[func]{backtrack}
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/* 回溯算法：N 皇后 */
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function backtrack(
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row: number,
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n: number,
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state: string[][],
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res: string[][][],
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cols: boolean[],
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diags1: boolean[],
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diags2: boolean[]
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): void {
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// 当放置完所有行时，记录解
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if (row === n) {
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res.push(state.map((row) => row.slice()));
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return;
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}
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// 遍历所有列
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for (let col = 0; col < n; col++) {
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// 计算该格子对应的主对角线和副对角线
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const diag1 = row - col + n - 1;
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const diag2 = row + col;
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// 剪枝：不允许该格子所在 (列 或 主对角线 或 副对角线) 包含皇后
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if (!(cols[col] || diags1[diag1] || diags2[diag2])) {
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// 尝试：将皇后放置在该格子
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state[row][col] = 'Q';
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cols[col] = diags1[diag1] = diags2[diag2] = true;
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// 放置下一行
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backtrack(row + 1, n, state, res, cols, diags1, diags2);
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// 回退：将该格子恢复为空位
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state[row][col] = '#';
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cols[col] = diags1[diag1] = diags2[diag2] = false;
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}
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}
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}
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[class]{}-[func]{nQueens}
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/* 求解 N 皇后 */
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function nQueens(n: number): string[][][] {
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// 初始化 n*n 大小的棋盘，其中 'Q' 代表皇后，'#' 代表空位
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const state = Array.from({ length: n }, () => Array(n).fill('#'));
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const cols = Array(n).fill(false); // 记录列是否有皇后
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const diags1 = Array(2 * n - 1).fill(false); // 记录主对角线是否有皇后
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const diags2 = Array(2 * n - 1).fill(false); // 记录副对角线是否有皇后
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const res: string[][][] = [];
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backtrack(0, n, state, res, cols, diags1, diags2);
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return res;
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}
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```
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=== "C"
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@@ -278,9 +433,49 @@ comments: true
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=== "Swift"
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```swift title="n_queens.swift"
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[class]{}-[func]{backtrack}
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/* 回溯算法：N 皇后 */
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func backtrack(row: Int, n: Int, state: inout [[String]], res: inout [[[String]]], cols: inout [Bool], diags1: inout [Bool], diags2: inout [Bool]) {
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// 当放置完所有行时，记录解
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if row == n {
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res.append(state)
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return
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}
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// 遍历所有列
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for col in 0 ..< n {
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// 计算该格子对应的主对角线和副对角线
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let diag1 = row - col + n - 1
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let diag2 = row + col
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// 剪枝：不允许该格子所在 (列 或 主对角线 或 副对角线) 包含皇后
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if !(cols[col] || diags1[diag1] || diags2[diag2]) {
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// 尝试：将皇后放置在该格子
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state[row][col] = "Q"
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cols[col] = true
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diags1[diag1] = true
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diags2[diag2] = true
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// 放置下一行
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backtrack(row: row + 1, n: n, state: &state, res: &res, cols: &cols, diags1: &diags1, diags2: &diags2)
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// 回退：将该格子恢复为空位
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state[row][col] = "#"
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cols[col] = false
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diags1[diag1] = false
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diags2[diag2] = false
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}
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}
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}
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[class]{}-[func]{nQueens}
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/* 求解 N 皇后 */
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func nQueens(n: Int) -> [[[String]]] {
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// 初始化 n*n 大小的棋盘，其中 'Q' 代表皇后，'#' 代表空位
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var state = Array(repeating: Array(repeating: "#", count: n), count: n)
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var cols = Array(repeating: false, count: n) // 记录列是否有皇后
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var diags1 = Array(repeating: false, count: 2 * n - 1) // 记录主对角线是否有皇后
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var diags2 = Array(repeating: false, count: 2 * n - 1) // 记录副对角线是否有皇后
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var res: [[[String]]] = []
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backtrack(row: 0, n: n, state: &state, res: &res, cols: &cols, diags1: &diags1, diags2: &diags2)
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return res
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}
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```
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=== "Zig"
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Reference in New Issue
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