lintcode Two Sum

Two Sum - Input array is sorted 

•  描述
•  笔记
•  数据
•  评测

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.

 注意事项

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

您在真实的面试中是否遇到过这个题？ 

Yes

样例

Given nums = [2, 7, 11, 15], target = 9
return [1, 2]

分析：做题的时候一定要读清楚题意！这道题题目中写明了find two numbers such that....，我还在考虑多个数字相加等于target的情况，真是可恨

下面上代码，第一种方法是达不到时间要求但是想法对的，参考过别人，因为用了二分法所以时间复杂度为O(nlgn)，差不多超过1000ms了

而方法二，13ms就搞定了  O(nlgn)和O(n)的差距也是大啊，下面是代码

方法一：

class Solution {//从头部开始，在后面用二分法查找符合条件的另一个数public:    /*     * @param nums an array of Integer     * @param target = nums[index1] + nums[index2]     * @return [index1 + 1, index2 + 1] (index1 < index2)     */    vector<int> twoSum(vector<int> &nums, int target) {        if(nums.size()<=0)return  vector<int>(-1);for(int i=0;i<nums.size();i++){int k=target-nums[i];int Left=i,Right=nums.size()-1;while(Left<=Right){int Mid=(Left+Right)/2;if(k==nums[Mid])return {i+1,Mid+1};else if(k<nums[Mid]){Right=Mid;}else if(k>nums[Mid])Left=Mid-1;}}    }};

方法二：用两个指针一前一后夹击

class Solution {public:    /*     * @param nums an array of Integer     * @param target = nums[index1] + nums[index2]     * @return [index1 + 1, index2 + 1] (index1 < index2)     */    vector<int> twoSum(vector<int> &nums, int target) {        if(nums.size()<=0)return  {-1};int s=0,e=nums.size()-1;//莫忘记减一int sum=nums[s]+nums[e];while(sum!=target){    if(sum<target)    {        s++;    }    else    {        e--;    }    sum=nums[s]+nums[e];}return{s+1,e+1};            }};