Code & Func

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:

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

然后是dicuss中的O(n^2)的解法：

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