611. Valid Triangle Number

Given an array consists of non-negative integers, your task is to count the number of triplets chosen from the array that can make triangles if we take them as side lengths of a triangle.

Example 1:

Input: [2,2,3,4]
Output: 3
Explanation:
Valid combinations are:
2,3,4 (using the first 2)
2,3,4 (using the second 2)
2,2,3

Note:

The length of the given array won't exceed 1000.
The integers in the given array are in the range of [0, 1000].

题目：给定一个数组，判断有多少种可以组成三角形的三元组。

思路：参考解析。首先三角形的条件是任意两边之和大于第三边，即a+b>c,a+c>b和c+b>a，有三个判断条件。直接暴力循环的话TLE。对数组nums进行排序后判断条件就可以缩减为1个a+b>c，原因在于排序后c>b，c>a，那么在不等式左侧加上一个正数，还是满足条件的。用first、second表示两个短边所在的序号，third表示长边所在的序号。那么针对nums[third]这个长边，两个短边肯定都在其左侧（因为nums排过序了），令first从序号0开始，second从序号third-1开始，那么我们可以不断的调整这两个短边直到他们相遇为主。调整的原则是，如果nums[first] + nums[second] > nums[third]，说明我们已经找到了一种三角形，此时因为固定second的位置，然后不断增加first，肯定都是满足上述的判断的，此时对应的三角形的个数为second - first，所以我们更新second的位置。

工程代码下载

class Solution {
public:
    int triangleNumber(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        int len = nums.size();
        int cnt = 0;
        for(int third = len - 1; third >= 2; --third){
            int first = 0;
            int second = third - 1;
            while(first < second){
                if(nums[first] + nums[second] > nums[third]){
                    //<first, second, third> such that first + second > third, there is no need to increment first
                    cnt += (second - first);
                    --second;
                }
                else
                    ++first;
            }
        }
        return cnt;
    }
};

同[LeetCode] 3Sum Smaller 三数之和较小值.