167. Two Sum II

Given an array of integers that is already sorted in ascending order, 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 and you may not use the same element twice.

Input: numbers={2, 7, 11, 15}, target=9
Output: index1=1, index2=2

tag：binary search

code:

class Solution {public:    vector<int> twoSum(vector<int>& numbers, int target) {        vector<int> ret;        int big;        int little;        int cur;        for(int i=0;i<numbers.size()-1;i++)        {            little=i;big=numbers.size()-1;            cur=big;            while(big-little>=1&&!ret.size())            {                if(numbers[i]+numbers[cur]<target&&cur==numbers.size()-1)                    break;                else if(numbers[i]+numbers[cur]<target)                {                    little=cur;                    cur=(little+big)/2;                }                else if(numbers[i]+numbers[cur]>target)                {                    big=cur;                    cur=(little+big)/2;                }                else                {                    ret.push_back(i+1);                    ret.push_back(cur+1);                }            }        }        return ret;    }};

存在的问题就是每次更新big little之后，可能停留在原先已经测试过的点，并没有前进，比如：

little=3，big=4，cur会一直停留在3，不会更新。

正确有效代码如下：

vector<int> twoSum(vector<int> &numbers, int target) {    if(numbers.empty()) return {};    for(int i=0; i<numbers.size()-1; i++) {        int start=i+1, end=numbers.size()-1, gap=target-numbers[i];        while(start <= end) {            int m = start+(end-start)/2;            if(numbers[m] == gap) return {i+1,m+1};            else if(numbers[m] > gap) end=m-1;            else start=m+1;        }    }}

更新的start和end都是已经检测过的m的左右的数值。