#443 Two Sum II

题目描述：

Given an array of integers, find how many pairs in the array such that their sum is bigger than a specific target number. Please return the number of pairs.

Have you met this question in a real interview? 

Yes

Example

Given numbers = [2, 7, 11, 15], target = 24. Return 1. (11 + 15 is the only pair)

Challenge 

Do it in O(1) extra space and O(nlogn) time.

题目思路：

把array从小到大进行sort，再用两个pointers：l指向array的头，r指向array的尾。当nums[l] + nums[r]的值满足条件时，这也表示l = l...r - 1和r的组合也满足条件，这时可以skip这一段的检查，下一个loop中r = r - 1；当他俩的值不满足条件时，这也表示r = l + 1...r和l的组合也不满足条件，这样也可以skip相应的情况。

Mycode（AC = 66ms）：

class Solution {public:    /**     * @param nums: an array of integer     * @param target: an integer     * @return: an integer     */    int twoSum2(vector<int> &nums, int target) {        // Write your code here        sort(nums.begin(), nums.end());                int l = 0, r = nums.size() - 1;        int num_pairs = 0;                while (l < r) {            // if sum > target, then             // sum of l = l...r-1 and r = r            // all > target            if (nums[l] + nums[r] > target) {                num_pairs += r - l;                r--;            }            // if sum <= target, then             // don't need to consider r = l + 1...r            else {                l++;            }        }                return num_pairs;    }};