【Two Sum】 --MyLeetCode(一)

Gang_Plus_1s

于 2017-08-19 12:36:20 发布

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分类专栏： myLeetcode 文章标签： Leetcode C++面试题

本文链接：https://blog.youkuaiyun.com/Gang_dada/article/details/77411158

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本文介绍了一种解决两数之和问题的经典算法。给定一个整数数组及目标值，寻找数组中两个数使它们的和等于目标值。文章提供了C++与C语言两种实现方式，通过遍历数组来找出满足条件的两个数及其下标。

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Description:

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].

思路：

类似冒泡排序，遍历数组，查找两个数字并返回下标：

自己实现的渣渣C++版：(129ms QAQ)

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        vector<int> res(2,0) ;
        for(int i = 0;i < nums.size();++i){
            for(int j = i + 1;j < nums.size();++j){
                if(nums[j] == target - nums[i] ){
                    res[0] = i;
                    res[1] = j;
                }
            }
        }
        return res;
    }
};

C语言版(106ms)：

/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
int* twoSum(int* nums, int numsSize, int target) {
    for(int i = 0;i < numsSize;++i){
        for( int j = i + 1;j < numsSize;++j){
            if( nums[i] + nums[j] == target){
                int* res = (int*)malloc(sizeof(int)*2);    
                *res = i;
                *(res + 1) = j;
                return res;
            }
        }
    }
    return 0;
}