1. Two Sum

标签：ecif   The   red   for   ==   not   sum   pre   class   

Given an array of integers, return indices of the two numbers such that they add up to a specific target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        for(int i=0; i<nums.size(); i++){
            for(int j=i+1; j<nums.size(); j++){
                if(nums[i] + nums[j] == target){
                    return {i,j};
                }
            }
        }
        return {};
    }
};

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        unordered_map<int,int> indies;
        for(int i=0; i<nums.size();++i)
            indies[nums[i]] = i;
        
        for(int i=0; i<nums.size(); ++i){
            int left = target - nums[i];
            if(indies.count(left) && indies[left] != i){
                return {i, indies[left]};
            }
        }
        return {};
    }
};

1. Two Sum

标签：ecif   The   red   for   ==   not   sum   pre   class   

原文地址：https://www.cnblogs.com/wtuwangxin/p/9751052.html