[旧版] LeetCode Two Sum

Given an array of integers, find two numbers such that they add up to a specific target number.

The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2. Please note that your returned answers (both index1 and index2) are not zero-based.

You may assume that each input would have exactly one solution.

Input: numbers={2, 7, 11, 15}, target=9

Output: index1=1, index2=2

class Solution {public:    vector<int> twoSum(vector<int> &numbers, int target) {        vector<int> v(numbers);        sort(v.begin(),v.end());        vector<int>::iterator ia = v.begin();        vector<int>::iterator ib = v.end();        int a,b;        ib--;        while(ia < ib) {            if(*ia+*ib > target) {                ib--;            }else if(*ia+*ib < target) {                ia++;            }else {                a = *ia;                b = *ib;                break;            }        }        vector<int> r;        for(int i = 0; i < numbers.size();i++) {            if(numbers[i]==a || numbers[i]==b) {                r.push_back(i+1);            }        }        return r;    }};

思路：暴力解法是O(n^2)。这种解法是O(nlogn)。

关键之处在于排序后首尾指针步进的解法可行，复杂度O(n)，算法复杂度为排序复杂度。

最后的for循环查找也很精妙。