Math-645. Set Mismatch - 相关文章

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645. Set Mismatch - LeetCode

Question 645. Set Mismatch Solution 思路: 遍历每个数字,然后将其应该出现的位置上的数字变为其相反数,这样如果我们再变为其相反数之前已经成负数了,说明该数字是重复数,将其将入结果res中,然后再遍历原数组,如果某个位置上的数字为正数,说明该位置对应的数字没有出现过,加入res中即可 Java实现: public int[] findErrorNums(int[] nums) { /* int a = 0; for (int i : nums) { if (nu…

448. Find All Numbers Disappeared in an Array&&645. Set Mismatch

题目: 448. Find All Numbers Disappeared in an Array 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…

【Leetcode_easy】645. Set Mismatch

problem 645. Set Mismatch 题意: solution1: 统计每个数字出现的次数了,然后再遍历次数数组,如果某个数字出现了两次就是重复数,如果出现了0次,就是缺失数. 注意:如何统计每个数字出现的次数: class Solution { public: vector<int> findErrorNums(vector<int>& nums) { vector<, ), cnt(nums.size(), ); ]; ; i<nums.siz…

645. Set Mismatch挑出不匹配的元素和应该真正存在的元素

［抄题］: he set S originally contains numbers from 1 to n. But unfortunately, due to the data error, one of the numbers in the set got duplicated to another number in the set, which results in repetition of one number and loss of another number. Given a…

LeetCode 645. Set Mismatch （集合不匹配）

The set S originally contains numbers from 1 to n. But unfortunately, due to the data error, one of the numbers in the set got duplicated to another number in the set, which results in repetition of one number and loss of another number. Given an arr…

645. Set Mismatch

static int wing=[]() { std::ios::sync_with_stdio(false); cin.tie(NULL); ; }(); class Solution { public: vector<int> findErrorNums(vector<int>& nums) { int sz=nums.size(); vector<,); for(int i:nums) help[i]++; help[]=; vector<int>…

leetcode 645. Set Mismatch——凡是要节约空间的题目都在输入数据上下功夫不要担心破坏原始的input