leetcode 167 Two Sum II - Input array is sorted

给定一个有序的数组，和一个整数目标，求两个数的和等于目标的索引，索引从1开始。假设必定存在解。

有两种思路：

直接找：

    vector<int> twoSum(vector<int>& numbers, int target) {int s = 0, e = numbers.size() - 1;while (true) {if (numbers[s] + numbers[e] > target)e--;else if (numbers[s] + numbers[e] < target)s++;else {return vector<int> {s + 1, e + 1};}}}

原理显而易见，最坏情况，也就O(n)，当两个索引位于中间位置。

二分查找：

    vector<int> twoSum(vector<int>& numbers, int target) {auto tail = upper_bound(numbers.begin(), numbers.end(), target);auto start = numbers.begin();for (;start != tail;) {int other = target - *start;++start;tail = lower_bound(start, tail, other);if (other == *tail) {vector<int> ret = {start - numbers.begin(), tail - numbers.begin() + 1};return move(ret);}other = target - *tail;--tail;start = lower_bound(start, tail, other);if (other == *start) {vector<int> ret = {start - numbers.begin() + 1, tail - numbers.begin() + 2};return move(ret);}}}

每次都是二分地前进后退，最坏情况下达到O(nlogn)，当两索引位于中间。

但当两个答案靠近某一端时性能较好。就leetcode结果来看，相差不大。

转载于:https://www.cnblogs.com/willaty/p/8376786.html

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leetcode 167 Two Sum II - Input array is sorted

有两种思路：

直接找：

二分查找：

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