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/* * 259. 3Sum Smaller * 2016-6-21 by Mingyang * Given an array of n integers nums and a target, * find the number of index triplets i, j, k with 0 <= i < j < k < n * that satisfy the condition nums[i] + nums[j] + nums[k] < target. * 这个题目就是翻版的3Sum,刚开始自己做逻辑有点乱 * 如果小于target直接count += right-left;因为第二个是left的话,第三个从left+1到right都是合法的 * 不用一个一个判断,另外,每次完了以后left++,另外若小于,自然是right-- */ public static int count; public static int threeSumSmaller(int[] nums, int target) { count = 0; Arrays.sort(nums); int len = nums.length; for(int i=0; i<len-2; i++) { int left = i+1, right = len-1; while(left < right) { if(nums[i] + nums[left] + nums[right] < target) { count += right-left; left++; } else { right--; } } } return count; }
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原文地址:http://www.cnblogs.com/zmyvszk/p/5606658.html