LeetCode 448. Find All Numbers Disappeared in an Array 442. Find All Duplicates in an Array
这两道题很有意思，由于元素为1~n，因此每个元素的值-1（映射到0~n-1）就可以直接当做下标。这样将 nums 中对应下标的元素 *-1 以i表示 index+1 这个元素出现过了，能节省存储的空间。
448. Find All Numbers Disappeared in an Array
class Solution {
public:vector<int> findDisappearedNumbers(vector<int>& nums) {for (int i=0;i<nums.size();++i){int index=abs(nums[i])-1;if (nums[index]>0) nums[index]*=-1;}vector<int> res;for (int i=0;i<nums.size();++i){if (nums[i]>0) res.push_back(i+1);}return res;}
};442. Find All Duplicates in an Array
class Solution {
public:vector<int> findDuplicates(vector<int>& nums) {vector<int> res;for (int i=0;i<nums.size();++i){int index=abs(nums[i])-1;if (nums[index]>0) nums[index]*=-1;else res.push_back(index+1);}return res;}
};转载于:https://www.cnblogs.com/hankunyan/p/9329164.html
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