LeetCode 题解(270) : 3Sum Smaller

题目：

Given an array of n integers nums and a target, find the number of index tripletsi, 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?

题解：

一步一步由O(n3)到O(n2)。

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

class Solution {
public:
    int threeSumSmaller(vector<int>& nums, int target) {
        sort(nums.begin(), nums.end());
        int k = nums.size() - 2;
        int count = 0;
        for(int i = 0; i < k; i++) {
            for(int j = i + 1; j < k + 1; j++) {
                int end = nums.size() - 1;
                int t = target - nums[i] - nums[j];
                while(end > j && nums[end] >= t)
                    end--;
                count += ((end > j) ? (end - j) : 0);
            }
        }
        return count;
    }
};

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