Problem
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]
]
Solution (Java)
class Solution {
public List<List<Integer>> combinationSum(int[] coins, int amount) {
List<List<Integer>> ans = new ArrayList<>();
List<Integer> subList = new ArrayList<>();
combinationSumRec(coins.length, coins, amount, subList, ans);
return ans;
}
private void combinationSumRec(
int n, int[] coins, int amount, List<Integer> subList, List<List<Integer>> ans) {
if (amount == 0 || n == 0) {
if (amount == 0) {
List<Integer> base = new ArrayList<>(subList);
ans.add(base);
}
return;
}
if (amount - coins[n - 1] >= 0) {
subList.add(coins[n - 1]);
combinationSumRec(n, coins, amount - coins[n - 1], subList, ans);
subList.remove(subList.size() - 1);
}
combinationSumRec(n - 1, coins, amount, subList, ans);
}
}
Solution (Javascript)
/**
* @param {number[]} candidates
* @param {number} target
* @return {number[][]}
*/
var combinationSum = function(candidates, target) {
var res = [];
var len = candidates.length;
candidates.sort((a, b) => (a - b));
dfs(res, [], 0, len, candidates, target);
return res;
};
var dfs = function (res, stack, index, len, candidates, target) {
var tmp = null;
if (target < 0) return;
if (target === 0) return res.push(stack);
for (var i = index; i < len; i++) {
if (candidates[i] > target) break;
tmp = Array.from(stack);
tmp.push(candidates[i]);
dfs(res, tmp, i, len, candidates, target - candidates[i]);
}
};
Explain:
对树进行深度优先的搜索。注意解法不能重复，即下次搜索从 index 开始
Complexity:
- Time complexity : O(n^2).
- Space complexity : O(n^2).