LeetCode Valid Triangle Number

· 184 words · 1 minute read
2017年6月
LeetCode

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

错解

刷题刷了一段时间了，凭借直觉，直接递归来求，发现在测试用例174时超时。先贴上超时代码.

public class Solution {
    int count = 0;
    public int triangleNumber(int[] nums) {
        find(nums,0,new ArrayList<Integer>());
        return count;
    }
    public void find(int[] nums, int pos, ArrayList<Integer> arrayList){
        if(arrayList.size() == 3){
            ArrayList<Integer> list = new ArrayList<>(arrayList);
            Collections.sort(list);
            if(list.get(0) + list.get(1) > list.get(2)){
                count++;
            }
            return;
        }
        for(int i = pos; i < nums.length; i++){
            arrayList.add(nums[i]);
            find(nums,i+1,arrayList);
            arrayList.remove(arrayList.size()-1);
        }
    }
}

正解

思路：为了判断是否满足三角形，需要判断两边之和是否大于第三边。先将整个数组排序，设定三个指针，l,r,p。从后往前遍历，p是末尾指针，r是末尾指针的前一位，l从0开始，若nums[l] + nums[r] > nums[i],则r-l范围的都满足条件，因为数组已经从小到大排序，所以若nums[l]符合条件，那么nums[l…r-1]也符合条件。若不符合nums[l] + nums[r] > nums[i]，则l++。时间复杂度为O(n^2)。

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