[Leetcode] 448. Find All Numbers Disappeared in an Array 解题报告

题目：

Given an array of integers where 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once.

Find all the elements of [1, n] inclusive that do not appear in this array.

Could you do it without extra space and in O(n) runtime? You may assume the returned list does not count as extra space.

Example:

Input:
[4,3,2,7,8,2,3,1]

Output:
[5,6]

思路：

将值为value的数置换到nums[value - 1]这个位置，该任务可以在O(n)的时间复杂度内完成。然后再遍历，一旦发现nums[i] != i + 1，说明该位置上的数缺失了。

代码：

class Solution {
public:
vector<int> findDisappearedNumbers(vector<int>& nums) {
for(int i = 0; i < nums.size(); ++i) {
while(nums[i] != i + 1 && nums[nums[i] - 1] != nums[i]) {
swap(nums[i], nums[nums[i] - 1]);
}
}
vector<int> ret;
for(int i = 0; i < nums.size(); ++i) {
if(nums[i] != i + 1) {
ret.push_back(i + 1);
}
}
return ret;
}
};