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40. 组合总和 II
给定一个候选人编号的集合 candidates 和一个目标数 target ,找出 candidates 中所有可以使数字和为 target 的组合。
candidates 中的每个数字在每个组合中只能使用 一次 。
注意:解集不能包含重复的组合。
示例 1:
输入: candidates = [10, 1, 2, 7, 6, 1, 5], target = 8,
输出:
[
[1, 1, 6],
[1, 2, 5],
[1, 7],
[2, 6]
]
示例 2:
输入: candidates = [2, 5, 2, 1, 2], target = 5,
输出:
[
[1, 2, 2],
[5]
]
提示:
1 <= candidates.length <= 100
1 <= candidates[i] <= 50
1 <= target <= 30
- 来源:力扣(LeetCode)
- 链接:https:eetcode.cn/problems/combination-sum-ii
javascript
/**
* @param {number[]} candidates
* @param {number} target
* @return {number[][]}
*/
var combinationSum2 = function (candidates, target) {
candidates.sort((a, b) => a - b)
const res = []
dfs(candidates, target, 0, [], 0, res)
return res
}
function dfs(candidates, target, start, combine, sum, res) {
if (sum >= target) {
if (sum == target) {
res.push(combine.slice())
}
return
}
for (let i = start; i < candidates.length; i++) {
if (i - 1 >= start && candidates[i - 1] == candidates[i]) {
continue
}
combine.push(candidates[i])
dfs(candidates, target, i + 1, combine, sum + candidates[i], res)
combine.pop()
}
}
console.log(combinationSum2([10, 1, 2, 7, 6, 1, 5], 8))
console.log(combinationSum2([2, 5, 1, 3], 5))
console.log(combinationSum2([5, 2, 1, 2], 5))