0039. Combination Sum

39. Combination Sum #

题目 #

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]
]

题目大意 #

给定一个无重复元素的数组 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 中的数字可以无限制重复被选取。

解题思路 #

  • 题目要求出总和为 sum 的所有组合,组合需要去重。
  • 这一题和第 47 题类似,只不过元素可以反复使用。

代码 #


package leetcode

import "sort"

func combinationSum(candidates []int, target int) [][]int {
	if len(candidates) == 0 {
		return [][]int{}
	}
	c, res := []int{}, [][]int{}
	sort.Ints(candidates)
	findcombinationSum(candidates, target, 0, c, &res)
	return res
}

func findcombinationSum(nums []int, target, index int, c []int, res *[][]int) {
	if target <= 0 {
		if target == 0 {
			b := make([]int, len(c))
			copy(b, c)
			*res = append(*res, b)
		}
		return
	}
	for i := index; i < len(nums); i++ {
		if nums[i] > target { // 这里可以剪枝优化
			break
		}
		c = append(c, nums[i])
		findcombinationSum(nums, target-nums[i], i, c, res) // 注意这里迭代的时候 index 依旧不变,因为一个元素可以取多次
		c = c[:len(c)-1]
	}
}


⬅️上一页

下一页➡️

Calendar Sep 6, 2020
Edit Edit this page
本站总访问量:  次 您是本站第  位访问者