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Backtracking - 40. Combination Sum II

Backtracking - 40. Combination Sum II

40. Combination Sum II

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]
]

思路:

与39题不同的是,数组中同一个元素只能在组合中出现一次,这里只需要先对数组排序,就能去重,做法就和39题一样了。

代码:

func combinationSum2(candidates []int, target int) [][]int {

    var res [][]int
    if candidates == nil || len(candidates) == 0 {
        return res
    }
    sort.Ints(candidates)
    dfs(&res, []int{}, candidates, target, 0)
    return res
}

func dfs(res *[][]int, temp []int, nums []int, target int, start int) {
    if target < 0 {
        return 
    }
    if target == 0 {
        *res = append(*res, append([]int{}, temp...))
    }
    
    for i := start; i < len(nums); i++ {
        if i != start && nums[i] == nums[i-1] {
            continue
        }
        temp = append(temp, nums[i]);
        dfs(res, temp, nums, target - nums[i], i+1)
        temp = temp[:len(temp)-1]
    }
}

标题:Backtracking - 40. Combination Sum II
作者:michael
地址:https://www.bitbo-liuyang.com/articles/2019/09/03/1567523992180.html

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