3Sum Smaller -- LeetCode

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.

For example, given nums = [-2, 0, 1, 3], and target = 2.

Return 2. Because there are two triplets which sums are less than 2:

[-2, 0, 1]

[-2, 0, 1]
[-2, 0, 3]

思路：将数组排序。枚举第一个数，假设它为第i个数，则triplet中的第二个数和第三个数则在数组[i+1, n-1]中。我们用两个指针left和right来找这两个数，向中间搜索。

当nums[i] + nums[left] + nums[right] < target时，right指针向左移动仍然会符合，因此这时候满足条件的结果数有right - left个，记下这个数值，然后将left向右移动一位；否则将right向左移动一位。最后返回所有的结果数之和。时间复杂度为O(N^2)。

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