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最简单的做法是通过swap把每个数字放到自己的位置上，比如5就放到idx = 4的地方，第一个不符合的位置就是缺的位置因为每个数字最多通过2次就能换到适当的位置上所以时间是O(n)，空间是O(1)也可以用union-find做，但是代码过于复杂

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