组合求和 Combination Sum

组合总和问题解析

最新推荐文章于 2025-09-12 17:31:31 发布

转载最新推荐文章于 2025-09-12 17:31:31 发布 · 94 阅读

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原文链接：https://my.oschina.net/liyurong/blog/1528471

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#数据结构与算法

为什么80%的码农都做不了架构师？>>>

问题：

Given a set of candidate numbers (C) (without duplicates) 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.
The solution set must not contain duplicate combinations.

For example, given candidate set [2, 3, 6, 7] and target 7,
A solution set is:

[
  [7],
  [2, 2, 3]
]

解决：

① 结果要求返回所有符合要求解的题十有八九都是要利用到递归，而且解题的思路都大同小异，需要另写一个递归函数。注意是可以连续选择同一个数加入组合的。与方法②一样

class Solution { //17 ms
public static List<List<Integer>> combinationSum(int[] candidates, int target){
Arrays.sort(candidates);
List<List<Integer>> res = new ArrayList<>();
List<Integer> cur = new ArrayList<>();
dfs(candidates,0,target,cur,res);
return res;
}
private static void dfs(int[] candidates, int i, int target, List<Integer> cur, List<List<Integer>> res) {//i表示当前遍历到的下标。
if(target < 0) return;
if(target == 0){
res.add(new ArrayList<>(cur));
return;
}
for(int j = i; j < candidates.length && target >= candidates[j]; j ++){
cur.add(candidates[j]);
dfs(candidates,j,target - candidates[j],cur,res);//递归向后查找
cur.remove(cur.size() - 1);//还原链表
}
}
}

② 简化

class Solution { //19ms
List<List<Integer>> res = new ArrayList<>();
public List<List<Integer>> combinationSum(int[] candidates, int target) {
List<Integer> cur = new ArrayList<>();
dfs(candidates, 0, target, cur);
return res;
}
private void dfs(int[] nums, int i,int target, List<Integer> cur) {//i表示当前遍历到的位置
if (target == 0) {
res.add(new ArrayList<>(cur));
// return;
}
for (int j = i; j < nums.length; j ++) {
if (nums[j] <= target) {
cur.add(nums[j]);
dfs(nums, j, target - nums[j], cur);
cur.remove(cur.size() - 1);
}
}
}
}

转载于:https://my.oschina.net/liyurong/blog/1528471