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.

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, 3]

Follow up:
Could you solve it in O(n2) runtime?

这道题跟3sum的区别就是数字可以重复使用……所以在代码里注释掉了重复性判断的语句，代码如下：

public class Solution {
    public int threeSumSmaller(int[] nums, int target) {
        Arrays.sort(nums);
        int res = 0;
        for (int i = 0; i < nums.length; i ++) {
            //if (i == 0 || nums[i] != nums[i - 1]) {
                int low = i + 1, high = nums.length - 1, sum = target - nums[i];
                while (low < high) {
                    if (nums[low] + nums[high] < sum) {
                        res += high - low;
                        low ++;
                        //while (low < high && nums[low] == nums[low - 1]) low ++;
                    } else {
                        high --;
                        while (low < high && nums[high] == nums[high + 1]) high --;
                    }
                }
            //}
        }
        return res;
    }
}