【LEETCODE】#1 两数之和（vector的用法）_leetcode两数之和vector-优快云博客

本文详细解析了经典的两数之和算法问题。针对给定整数数组及目标值，找出数组中两个数的位置使得它们相加等于目标值。提供C++与C两种实现方式，帮助读者理解算法原理及不同语言的实现细节。

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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].

C++：

1、C++ vector的用法：http://www.cnblogs.com/Nonono-nw/p/3462183.html

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

【2019/3/14】

上次写这道题，还是去年的六月份，现在已经快一年过去了，leetcode中文官网出来了……

从C换成C++，重上刷题路，心里五味杂陈

C：

/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
int* twoSum(int* nums, int numsSize, int target) {
    int *s;
    s=(int*)malloc(2*sizeof(int));
    int i,j;
    for(i=0;i<numsSize-1;i++){    //注意上下限
        for(j=i+1;j<numsSize;j++){    //注意上下限
            if((*(nums+i)+*(nums+j))==target){
                *s=i;*(s+1)=j;
                return s;
            }
        }
    }
    return NULL;
}