[LeetCode] 39. Combination Sum ☆☆☆(数组相加等于指定的数)
https://leetcode.wang/leetCode-39-Combination-Sum.html
描述
Given a set of candidate numbers (candidates) (without duplicates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.
The same repeated number may be chosen from candidates unlimited number of times.
Note:
All numbers (including target) will be positive integers.
The solution set must not contain duplicate combinations.
Example 1:
Input: candidates = [2,3,6,7], target = 7,
A solution set is:
[
[7],
[2,2,3]
]
Example 2:
Input: candidates = [2,3,5], target = 8,
A solution set is:
[
[2,2,2,2],
[2,3,3],
[3,5]
]
解析
回溯法
参考这里) ,就是先向前列举所有情况,得到一个解或者走不通的时候就回溯。和37题有异曲同工之处,也算是回溯法很典型的应用,直接看代码吧。
动态规划
看原文
代码
回溯法
public List<List<Integer>> combinationSum(int[] nums, int target) {
List<List<Integer>> list = new ArrayList<>();
backtrack(list, new ArrayList<>(), nums, target, 0);
return list;
} private void backtrack(List<List<Integer>> list, List<Integer> tempList, int [] nums, int remain, int start){
if(remain < 0) return;
else if(remain == 0) list.add(new ArrayList<>(tempList));
else{
for(int i = start; i < nums.length; i++){
tempList.add(nums[i]);
backtrack(list, tempList, nums, remain - nums[i], i);
//找到了一个解或者 remain < 0 了,将当前数字移除,然后继续尝试
tempList.remove(tempList.size() - 1);
}
}
}
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