Valid-Triangle-Number

第67天。

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].

莫名其妙的用一个O(n^3)的解法AC了:

int triangleNumber(vector<int>& nums) {
    sort(nums.begin(),nums.end());
    int ret = 0;
    for(int i = 0;i < nums.size();i++) {
        for(int j = i + 1;j < nums.size();j++) {
            //nums[i] + num[j] > a;
            for(int k = j+1;k < nums.size() && nums[i] + nums[j] > nums[k];k++)
                ret++;
        }
    }
    return ret;
}

然后是dicuss中的O(n^2)的解法：

public static int triangleNumber(int[] A) {
    Arrays.sort(A);
    int count = 0, n = A.length;
    for (int i=n-1;i>=2;i--) {
        int l = 0, r = i-1;
        while (l < r) {
            if (A[l] + A[r] > A[i]) {
                count += r-l;
                r--;
            }
            else l++;
        }
    }
    return count;
}