259. 3Sum Smaller

Given an array of n integers nums and a target, find the number of index triplets i, j, k with 0 <= i < j < k < n that satisfy the condition nums[i] + nums[j] + nums[k] < target.

Example:

Input: nums = [-2,0,1,3], and target = 2
Output: 2
Explanation: Because there are two triplets which sums are less than 2:
[-2,0,1]
[-2,0,3]

Follow up:

Could you solve it in O(n2) runtime?

方法1: two pointers

grandyang：https://www.cnblogs.com/grandyang/p/5235086.html

思路：

仍然是3sum的方法，先确定第一个数，在[0, n - 3]的范围内遍历。从两端开始收缩双指针，这时会遇到两种情况：1. 3sum已经大于等于target，那么由于我们想要求的是smaller than target的组合，此时即使再移动左指针只会越来越大，肯定不会有结果，所以需要将右指针–。2. 小于target，此时找到了一个解，并且此时可以一次性累加所有right - left之间的结果，这是因为如果i, left, right 满足条件，那么所有i, left, [left + 1, right]形式的组合都能满足。累加之后，可以将left++，再次尝试此时的right是否过大。

Complexity

Time complexity: O(n^2)
Space complexity: O(1)

class Solution {
public:
    int threeSumSmaller(vector<int>& nums, int target) {
        int n = nums.size(), res = 0;
        sort(nums.begin(), nums.end());
        for (int i = 0; i < n - 2; i++) {
            int left = i + 1, right = n - 1;
            while (left < right) {
                if (nums[i] + nums[left] + nums[right] < target) {
                    res += right - left;
                    left++;
                } 
                else {
                    right--;
                }
            }
        }
        return res;
    }
};