[leetcode 1] Two Sum

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Question:

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.

Example:

Given nums = [2, 7, 11, 15], target = 9,Because nums[0] + nums[1] = 2 + 7 = 9,return [0, 1].

分析:

暴力法很简单，二重循环，但是超时了；

可以声明一个辅助数组，是对原数组的拷贝，这样，可以将辅助数组进行排序，排序后通过左右两个索引指针就可以判断是哪两个值相加为target；

然后在原数组中找到这两个值所对应的下标。

代码如下：

<span style="background-color: rgb(255, 255, 255);"><span style="font-size:14px;">class Solution {public:    vector<int> twoSum(vector<int>& nums, int target) {        vector<int> res(2);        vector<int> cp(nums);        sort(nums.begin(),nums.end());        int left = 0;        int right = nums.size()-1;        while(left < right){            int sum = nums[left] + nums[right];            if(sum < target)                ++left;            else if(sum > target)                --right;            else{                break;            }        }        for(int i = 0; i < nums.size(); ++i){            if(cp[i] == nums[left]){                left = i;                break;            }        }        for(int i = 0; i < nums.size(); ++i){            if(cp[i] == nums[right] && i != left){                right = i;                break;            }        }        res[0] = min(left,right);        res[1] = max(left,right);        return res;                 //暴力法       /* for(int i = 0; i < nums.size()-1; ++i){            for(int j = i+1; j < nums.size(); ++j){                if(nums[i] + nums[j] == target){                    res.push_back(i);                    res.push_back(j);                    break;                }            }        }        return res;*/    }};</span></span>

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