相似的代码，相同的功能，看不出有什么不同

Question

https://leetcode.com/problems/find-all-numbers-dispapered-in-an-array/discuss/93007/simple-java-in-place-sort-solution

你能查一下上面的链接吗？

我看不懂密码

1) nums[i] != i+1
2) nums[i] != nums[nums[i]-1]

index 0 : 1
index 1 : 2
index 2 : 3

然后，第一个只是简单地使用index我们可以检查index+1是不是值。

第二个,

nums[0] = nums[nums[i]-1]
nums[0] = nums[nums[0]-1]
nums[0] = nums[1-1]
nums[0] = nums[0]

它最终也是一样的东西，只是为了证明索引值=index+1。

但是为什么while循环必须同时具备这两个条件呢？或者我们就用一个？

Answer 1

1. 数字[i]！=i+1

值在它的位置吗？如果不是，可以把它换回原处...

这是需要的，因为你必须测试每一个位置

这是需要的，因为算法将链中的每个元素放置到其位置。

举个例子：

[3,1,2,4,6,5,8,7]

应该很清楚，你需要重新排列1，2，3和5，6和7，8。

让我们看看排序是如何发生的：

i:0 [3,1,2,4,6,5,8,7] 3<->2
i:0 [2,1,3,4,6,5,8,7] 2<->1
i:0 [1,2,3,4,6,5,8,7] now 1 is at its place, go to the right and find another chain
i:1 [1,2,3,4,6,5,8,7] 2 at its place
i:2 [1,2,3,4,6,5,8,7] 3 at its place
i:3 [1,2,3,4,6,5,8,7] 4 at its place
i:4 [1,2,3,4,6,5,8,7] 6<->5
i:4 [1,2,3,4,5,6,8,7] now is 5 at its place, go to the right and find another chain
i:5 [1,2,3,4,5,6,8,7] 6 at its place
i:6 [1,2,3,4,5,6,8,7] 8<->7
i:6 [1,2,3,4,5,6,7,8] now is 7 at its place, go to the right and find another chain
i:7 [1,2,3,4,5,6,7,8] 8 at its place
END