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Update the captions of all the figures.
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2. 同理,对剩余 $n - 1$ 个元素执行「冒泡」,可将第二大元素交换至正确位置,因而待排序元素只剩 $n - 2$ 个。
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3. 以此类推…… **循环 $n - 1$ 轮「冒泡」,即可完成整个数组的排序**。
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<p align="center"> Fig. 冒泡排序流程 </p>
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=== "Java"
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