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  • 611. Valid Triangle Number (solution 2)

    package LeetCode_611
    
    /**
     * 611. Valid Triangle Number
     * https://leetcode.com/problems/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].
     * */
    class Solution {
        /*
        Because Triangle is the sum of any two sides must be greater than third sides (任意两条边之和要大于第三边),
        we need to find 3 numbers,i<j<k and nums[i]+nums[j]>nums[k];
          Solution 1: sort array, bruce force, Time:O(n^3), Space:O(1);
        * Solution 2: sort array, two pointer, Time:O(n^2), Space:O(1);
        * */
        fun triangleNumber(nums: IntArray): Int {
            if (nums == null || nums.isEmpty() || nums.size < 2) {
                return 0
            }
            var count = 0
            nums.sort()
            //solution 2
            for (i in nums.indices) {
                var left = 0
                var right = i - 1
                while (left < right) {
                    if (nums[left] + nums[right] > nums[i]) {
                        /*
                       * for example array is :1,2,.,..5; if 1,2,5 is valid triplets, because array is sorted,
                       * so the element at the left of 5(right pointer) and at the right of 2(left pointer) are all valid.
                       * */
                        count += right - left
                        right--
                    } else {
                        left++
                    }
                }
            }
            return count
        }
    }
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  • 原文地址:https://www.cnblogs.com/johnnyzhao/p/14060670.html
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