Combination Sum | && || Leetcode
2016-03-24 21:10
501 查看
https://leetcode.com/problems/combination-sum/
Given a set of candidate numbers (C) and a target number (T), find all unique combinations in C where
the candidate numbers sums to T.
The same repeated number may be chosen from C unlimited number of times.
Note:
All numbers (including target) will be positive integers.
Elements in a combination (a1, a2,
… , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤
… ≤ ak).
The solution set must not contain duplicate combinations.
For example, given candidate set
A solution set is:
这个题目其实就是经典回溯问题,但是就是有一个不太一样地方就是,这里每个数可以出现无限制次数,加入说每个数只出现一次就跟这道题一样啦:
http://blog.csdn.net/huruzun/article/details/21823343
递归调用时需要增加一个控制起始位置参数来使得每次还可以调用出现过的数
接着是这道题的变形
https://leetcode.com/problems/combination-sum-ii/
与前面区别就是每个数字只能使用一次,而且要避免重复的答案,最简单就是在最终答案处进行判重,但是这样效率不高,在对数组排序后,如果某个数字后面又重复的就需要跳过。
public class Solution {
public static List<List<Integer>> combinationSum2(int[] candidates,
int target) {
List<List<Integer>> result = new ArrayList<>();
if (candidates == null || candidates.length == 0)
return result;
ArrayList<Integer> current = new ArrayList<Integer>();
Arrays.sort(candidates);
combinationSum(candidates, target, 0, current, result);
return result;
}
public static void combinationSum(int[] candidates, int target, int i,
ArrayList<Integer> curr, List<List<Integer>> result) {
if (target == 0) {
// 这里一定要new起来一个新的拷贝
ArrayList<Integer> temp = new ArrayList<Integer>(curr);
result.add(temp);
return;
}
for (; i < candidates.length; i++) {
if (target < candidates[i])
return;
curr.add(candidates[i]);
combinationSum(candidates, target - candidates[i], i+1, curr, result);
curr.remove(curr.size() - 1);
// 这里重复的数字要跳过
while(i< candidates.length-1 && candidates[i] == candidates[i+1]){
i++;
}
}
}
}
Given a set of candidate numbers (C) and a target number (T), find all unique combinations in C where
the candidate numbers sums to T.
The same repeated number may be chosen from C unlimited number of times.
Note:
All numbers (including target) will be positive integers.
Elements in a combination (a1, a2,
… , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤
… ≤ ak).
The solution set must not contain duplicate combinations.
For example, given candidate set
2,3,6,7and target
7,
A solution set is:
[7]
[2, 2, 3]
这个题目其实就是经典回溯问题,但是就是有一个不太一样地方就是,这里每个数可以出现无限制次数,加入说每个数只出现一次就跟这道题一样啦:
http://blog.csdn.net/huruzun/article/details/21823343
递归调用时需要增加一个控制起始位置参数来使得每次还可以调用出现过的数
package combination_Sum; import java.util.ArrayList; import java.util.Arrays; import java.util.List; public class Solution { public static List<List<Integer>> combinationSum(int[] candidates, int target) { List<List<Integer>> result = new ArrayList<>(); if(candidates == null || candidates.length == 0) return result; ArrayList<Integer> current = new ArrayList<Integer>(); Arrays.sort(candidates); combinationSum(candidates, target, 0, current, result); return result; } public static void combinationSum(int[] candidates, int target, int j, ArrayList<Integer> curr, List<List<Integer>> result){ if(target == 0){ // 这里一定要new起来一个新的拷贝 ArrayList<Integer> temp = new ArrayList<Integer>(curr); result.add(temp); return; } // 这里 参数 j 用来控制同一个数可以出现多次 for(int i=j; i<candidates.length; i++){ if(target < candidates[i]) return; curr.add(candidates[i]); combinationSum(candidates, target - candidates[i], i, curr, result); curr.remove(curr.size()-1); } } public static void main(String[] args){ int []candidates = {2,3,6,7}; System.out.println(combinationSum(candidates,7)); } }
接着是这道题的变形
https://leetcode.com/problems/combination-sum-ii/
与前面区别就是每个数字只能使用一次,而且要避免重复的答案,最简单就是在最终答案处进行判重,但是这样效率不高,在对数组排序后,如果某个数字后面又重复的就需要跳过。
public class Solution {
public static List<List<Integer>> combinationSum2(int[] candidates,
int target) {
List<List<Integer>> result = new ArrayList<>();
if (candidates == null || candidates.length == 0)
return result;
ArrayList<Integer> current = new ArrayList<Integer>();
Arrays.sort(candidates);
combinationSum(candidates, target, 0, current, result);
return result;
}
public static void combinationSum(int[] candidates, int target, int i,
ArrayList<Integer> curr, List<List<Integer>> result) {
if (target == 0) {
// 这里一定要new起来一个新的拷贝
ArrayList<Integer> temp = new ArrayList<Integer>(curr);
result.add(temp);
return;
}
for (; i < candidates.length; i++) {
if (target < candidates[i])
return;
curr.add(candidates[i]);
combinationSum(candidates, target - candidates[i], i+1, curr, result);
curr.remove(curr.size() - 1);
// 这里重复的数字要跳过
while(i< candidates.length-1 && candidates[i] == candidates[i+1]){
i++;
}
}
}
}
相关文章推荐
- 《世界是数字的》读后感
- Sort function 应用
- COCOS学习笔记--Button类及其相关控件属性
- Python实现的淘宝直通车数据抓取(4)
- spring的两种声眀式事务
- ubuntu12.04普通用户登陆时,忘记密码??
- 2.0-Redis配置讲解(上)
- linux下文件类型获取
- Python调用SIFT出现的问题:No such file or directory: '*.sift' 解决办法,超好用
- 数据类型、变量、常量
- [DB2, Derby, H2, HSQL, Informix, MS-SQL, MySQL, Oracle, PostgreSQL, Sybase]
- 1006-Elevator-解题报告
- tinyxml源代码解析(一)
- Java 学习笔记(一)
- NVIDIA显卡驱动未加载问题——未完全解决
- ffmpeg的时间戳
- 软金测试第一次作业
- 静态、自适应、流式、响应式四种网页布局的区别
- hdu1054 树形dp&&二分图
- Microservice Anti-patterns