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题解一源码分析对于`i >= right`的讨论，认为没有必要。因为`for 循环`保证了 `i` 每次都增加`1`，而 `right` 是满足条件才增加`1`，所以`i >= right`恒成立，只需要 `if (nums[i] < k)` 就够了。 另外，`nums[i]` 与 `nums[right]` 的 `swap` 操作，可以加上 `if (i != right)` 限定，避免了自己与自己的交换，代码也更好理解～
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题解一源码分析对于
i >= right的讨论，认为没有必要。因为for 循环保证了i每次都增加1，而right是满足条件才增加1，所以i >= right恒成立，只需要if (nums[i] < k)就够了。另外，
nums[i]与nums[right]的swap操作，可以加上if (i != right)限定，避免了自己与自己的交换，代码也更好理解～