Skip to content

39. 组合总和 #30

@Geekhyt

Description

@Geekhyt

原题链接

回溯

先明确,元素可以重复使用,但是组合不能重复。

  1. 使用回溯法,不符合条件的情况进行剪枝。
  2. cur === target 时,拷贝 arr 推进结果集。
  3. 从 start 开始遍历可选数组,选择当前数字后递归时注意要基于当前状态 i 继续选择,因为元素是可以重复进入集合的。
  4. 撤销选择,恢复状态。
const combinationSum = (candidates, target) => {
    const res = []
    // start: 起点索引 arr: 当前集合 cur: 当前所求之和
    const dfs = (start, arr, cur) => {
        if (cur > target) return
        if (cur === target) {
            res.push(arr.slice())
            return
        }
        for (let i = start; i < candidates.length; i++) {
            arr.push(candidates[i])
            dfs(i, arr, cur + candidates[i])
            arr.pop()
        }
    }
    dfs(0, [], 0)
    return res
}

Metadata

Metadata

Assignees

No one assigned

    Labels

    Projects

    No projects

    Milestone

    No milestone

    Relationships

    None yet

    Development

    No branches or pull requests

    Issue actions

      pFad - Phonifier reborn

      Pfad - The Proxy pFad of © 2024 Garber Painting. All rights reserved.

      Note: This service is not intended for secure transactions such as banking, social media, email, or purchasing. Use at your own risk. We assume no liability whatsoever for broken pages.


      Alternative Proxies:

      Alternative Proxy

      pFad Proxy

      pFad v3 Proxy

      pFad v4 Proxy