0040. Combination Sum I I

# 40. Combination Sum II#

## 题目 #

Given a collection of candidate numbers (`candidates`) and a target number (`target`), find all unique combinations in `candidates` where the candidate numbers sums to `target`.

Each number in `candidates` may only be used once in the combination.

Note:

• All numbers (including `target`) will be positive integers.
• The solution set must not contain duplicate combinations.

Example 1:

``````Input: candidates = [10,1,2,7,6,1,5], target = 8,
A solution set is:
[
[1, 7],
[1, 2, 5],
[2, 6],
[1, 1, 6]
]
``````

Example 2:

``````Input: candidates = [2,5,2,1,2], target = 5,
A solution set is:
[
[1,2,2],
[5]
]
``````

## 题目大意 #

candidates 中的每个数字在每个组合中只能使用一次。

## 解题思路 #

• 题目要求出总和为 sum 的所有组合，组合需要去重。这一题是第 39 题的加强版，第 39 题中元素可以重复利用(重复元素可无限次使用)，这一题中元素只能有限次数的利用，因为存在重复元素，并且每个元素只能用一次(重复元素只能使用有限次)
• 这一题和第 47 题类似。

## 代码 #

``````
package leetcode

import (
"sort"
)

func combinationSum2(candidates []int, target int) [][]int {
if len(candidates) == 0 {
return [][]int{}
}
c, res := []int{}, [][]int{}
sort.Ints(candidates) // 这里是去重的关键逻辑
findcombinationSum2(candidates, target, 0, c, &res)
return res
}

func findcombinationSum2(nums []int, target, index int, c []int, res *[][]int) {
if target == 0 {
b := make([]int, len(c))
copy(b, c)
*res = append(*res, b)
return
}
for i := index; i < len(nums); i++ {
if i > index && nums[i] == nums[i-1] { // 这里是去重的关键逻辑,本次不取重复数字，下次循环可能会取重复数字
continue
}
if target >= nums[i] {
c = append(c, nums[i])
findcombinationSum2(nums, target-nums[i], i+1, c, res)
c = c[:len(c)-1]
}
}
}

``````

Apr 8, 2023