标签:target temp nums int backtrace result Beaconfire todo screening
https://drive.google.com/drive/folders/1BTKV760sLnrdtP8WhGJQaqpZjoaEZhUd?usp=sharing
如何避免kafka中的一个消息被消耗两次 -分区
hashmap内部的结构是什么样的,两个值相同的object当作key size会是多少
回溯法初始的index = 0,才能算出重复值 eg (1,3) (3,1)
复杂度:数组元素个数n的(target)次方
/** Given an array of distinct integers nums and a target integer target, return the number of possible combinations that add up to target. Input: nums = [1,2,3], target = 4 Output: 7 Explanation: The possible combination ways are: (1, 1, 1, 1) (1, 1, 2) (1, 2, 1) (1, 3) (2, 1, 1) (2, 2) (3, 1) Note that different sequences are counted as different combinations. result = [[1, 1, 1, 1], [1, 1, 2], [1, 3], [2, 2]] */ import java.util.*; public class MyClass { public static void main(String args[]) { int[] nums = {1,2,3}; int target = 4; List<List<Integer>> result = new ArrayList<List<Integer>>(); //sort Arrays.sort(nums); backtrace(nums, 0, new ArrayList<>(), result, 0, target); System.out.println("result = " + result); System.out.println("result.size() = " + result.size()); } public static void backtrace(int[] nums, int start, List<Integer> temp, List<List<Integer>> result, int currSum, int target) { //exit case if (currSum == target) { result.add(new ArrayList<>(temp)); }else if (currSum > target) { return ; }else { for (int i = 0; i < nums.length; i++) { //handle duplicate // if (temp.contains(nums[i])) // continue; //backtrace temp.add(nums[i]); backtrace(nums, i, temp, result, currSum + nums[i], target); temp.remove(temp.size() - 1); } } } }View Code
标签:target,temp,nums,int,backtrace,result,Beaconfire,todo,screening 来源: https://www.cnblogs.com/immiao0319/p/15328277.html
本站声明: 1. iCode9 技术分享网(下文简称本站)提供的所有内容,仅供技术学习、探讨和分享; 2. 关于本站的所有留言、评论、转载及引用,纯属内容发起人的个人观点,与本站观点和立场无关; 3. 关于本站的所有言论和文字,纯属内容发起人的个人观点,与本站观点和立场无关; 4. 本站文章均是网友提供,不完全保证技术分享内容的完整性、准确性、时效性、风险性和版权归属;如您发现该文章侵犯了您的权益,可联系我们第一时间进行删除; 5. 本站为非盈利性的个人网站,所有内容不会用来进行牟利,也不会利用任何形式的广告来间接获益,纯粹是为了广大技术爱好者提供技术内容和技术思想的分享性交流网站。