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Backtracking - 39. Combination Sum

Backtracking - 39. Combination Sum

39. Combination Sum

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]
]

思路:

题目意思是找出数组中所有能组合成target的组合,典型的回溯题目,递归枚举所有结果即可。

代码:

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

    var res [][]int

    if candidates == nil || len(candidates) == 0 {
        return res
    }
    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++ {
        temp = append(temp, nums[i]);
        dfs(res, temp, nums, target - nums[i], i)
        temp = temp[:len(temp)-1]
    }
}

标题:Backtracking - 39. Combination Sum
作者:michael
地址:https://www.bitbo-liuyang.com/articles/2019/09/03/1567522662605.html

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