LeetCode 448. Find All Numbers Disappeared in an Array & 442. Find All Duplicates in an Array

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這兩道題很有意思，由於元素為1~n，因此每個元素的值-1（映射到0~n-1）就可以直接當做下標。這樣將 nums 中對應下標的元素 *-1 以i表示 index+1 這個元素出現過了，能節省儲存的空間。

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;    }};

 

LeetCode 448. Find All Numbers Disappeared in an Array & 442. Find All Duplicates in an Array