259. 3Sum Smaller

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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 conditionnums[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?

这道题的解法与所有的3 sum，4 sum一样。唯一的区别在于两边往中间找的时候，如果小于dif，那么中间所有的都满足小于的情况，比如[3,1,0,-2], 4。 Sort完是[-2,0,1,3], 一开始选-2时，dif为6，left为1，right为3，小于dif，则以right为一个，left从左一直到right-1的组合都是满足要求的。

public int ThreeSumSmaller(int[] nums, int target) {
        Array.Sort(nums);
        var res =0;
        int size = nums.Count();
        for(int i =0;i< size;i++)
        {
            int num = nums[i];
            int dif = target - num;
            int left =i+1;
            int right = size -1;
            while(left<right)
            {
                if(nums[left]+nums[right] < dif)
                {
                    res+=right-left;
                    left++;
                }
                else right--;
            }
        }
        return res;
    }

259. 3Sum Smaller

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原文地址：http://www.cnblogs.com/renyualbert/p/5907839.html