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: 3Explanation:Valid combinations are: 2,3,4 (using the first 2)2,3,4 (using the second 2)2,2,3

Note:

    1. The length of the given array won't exceed 1000.    2. The integers in the given array are in the range of [0, 1000].

先给出自己的未能AC的代码，一般思路，回溯法，超时，意料之内，但还是写了出来。

bool isTriangle(vector<int>& path){    int a = path[0], b = path[1], c = path[2];    if (a + b > c && a + c > b && b + c > a)return true;    else return false;}void dfs_Tri(vector<int>& nums, int pos, vector<int>& path, int &cnt){    if (path.size() == 3){        if (isTriangle(path)){            cnt++;          }        return;    }    for (int i = pos; i < nums.size(); i++){        path.push_back(nums[i]);        dfs_Tri(nums, i + 1, path, cnt);        path.pop_back();    }}int triangleNumber(vector<int>& nums) {    if (nums.size() < 3)return 0;    vector<int> path;    int cnt = 0;    dfs_Tri(nums, 0, path, cnt);    return cnt;}

其实此题要用一个O（N^2 * log N）去解决，先对内部的元素排序，然后找出对小于两个元素的起始的内部元素的坐标，然后每次循环相加，找出小于两边之和的方法用二分查找法；

int triangleNumber(vector<int>& nums) {    if (nums.size() < 3)return 0;    sort(nums.begin(), nums.end());    int low, mid, high;    int cnt = 0;    for (int i = 0; i < nums.size() - 2; i++){        for (int j = i + 1; j < nums.size() - 1; j++){            int sum = nums[i] + nums[j];            low = j + 1, high = nums.size() - 1;            while (low <= high){                mid = (low + high) / 2;                if (nums[mid] >= sum)high = mid - 1;                else low = mid + 1;            }            cnt += low - j - 1;        }    }    return cnt;}