From 7284dffc429b87b75ceae98638cbc186544b28b1 Mon Sep 17 00:00:00 2001 From: Yuhao Ju Date: Thu, 8 Dec 2022 18:06:21 +0800 Subject: [PATCH] =?UTF-8?q?update=200040.=E7=BB=84=E5=90=88=E6=80=BB?= =?UTF-8?q?=E5=92=8CII:=20=E4=BF=AE=E6=94=B9=E9=94=99=E5=AD=97=EF=BC=8C?= =?UTF-8?q?=E6=9B=B4=E6=8D=A2=20go=20=E4=BB=A3=E7=A0=81?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0040.组合总和II.md | 108 ++++++++++++++++---------------- 1 file changed, 53 insertions(+), 55 deletions(-) diff --git a/problems/0040.组合总和II.md b/problems/0040.组合总和II.md index fcfc396b..2d4b836a 100644 --- a/problems/0040.组合总和II.md +++ b/problems/0040.组合总和II.md @@ -110,13 +110,13 @@ if (sum == target) { } ``` -`sum > target` 这个条件其实可以省略,因为和在递归单层遍历的时候,会有剪枝的操作,下面会介绍到。 +`sum > target` 这个条件其实可以省略,因为在递归单层遍历的时候,会有剪枝的操作,下面会介绍到。 * **单层搜索的逻辑** 这里与[39.组合总和](https://programmercarl.com/0039.组合总和.html)最大的不同就是要去重了。 -前面我们提到:要去重的是“同一树层上的使用过”,如果判断同一树层上元素(相同的元素)是否使用过了呢。 +前面我们提到:要去重的是“同一树层上的使用过”,如何判断同一树层上元素(相同的元素)是否使用过了呢。 **如果`candidates[i] == candidates[i - 1]` 并且 `used[i - 1] == false`,就说明:前一个树枝,使用了candidates[i - 1],也就是说同一树层使用过candidates[i - 1]**。 @@ -438,76 +438,74 @@ class Solution: **使用used数组** ```go +var ( + res [][]int + path []int + used []bool +) func combinationSum2(candidates []int, target int) [][]int { - var trcak []int - var res [][]int - var history map[int]bool - history=make(map[int]bool) - sort.Ints(candidates) - backtracking(0,0,target,candidates,trcak,&res,history) + res, path = make([][]int, 0), make([]int, 0, len(candidates)) + used = make([]bool, len(candidates)) + sort.Ints(candidates) // 排序,为剪枝做准备 + dfs(candidates, 0, target) return res } -func backtracking(startIndex,sum,target int,candidates,trcak []int,res *[][]int,history map[int]bool){ - //终止条件 - if sum==target{ - tmp:=make([]int,len(trcak)) - copy(tmp,trcak)//拷贝 - *res=append(*res,tmp)//放入结果集 + +func dfs(candidates []int, start int, target int) { + if target == 0 { // target 不断减小,如果为0说明达到了目标值 + tmp := make([]int, len(path)) + copy(tmp, path) + res = append(res, tmp) return } - if sum>target{return} - //回溯 - // used[i - 1] == true,说明同一树枝candidates[i - 1]使用过 - // used[i - 1] == false,说明同一树层candidates[i - 1]使用过 - for i:=startIndex;i0&&candidates[i]==candidates[i-1]&&history[i-1]==false{ - continue + for i := start; i < len(candidates); i++ { + if candidates[i] > target { // 剪枝,提前返回 + break } - //更新路径集合和sum - trcak=append(trcak,candidates[i]) - sum+=candidates[i] - history[i]=true - //递归 - backtracking(i+1,sum,target,candidates,trcak,res,history) - //回溯 - trcak=trcak[:len(trcak)-1] - sum-=candidates[i] - history[i]=false + // used[i - 1] == true,说明同一树枝candidates[i - 1]使用过 + // used[i - 1] == false,说明同一树层candidates[i - 1]使用过 + if i > 0 && candidates[i] == candidates[i-1] && used[i-1] == false { + continue + } + path = append(path, candidates[i]) + used[i] = true + dfs(candidates, i+1, target - candidates[i]) + used[i] = false + path = path[:len(path) - 1] } } ``` **不使用used数组** ```go +var ( + res [][]int + path []int +) func combinationSum2(candidates []int, target int) [][]int { - var trcak []int - var res [][]int - sort.Ints(candidates) - backtracking(0,0,target,candidates,trcak,&res) + res, path = make([][]int, 0), make([]int, 0, len(candidates)) + sort.Ints(candidates) // 排序,为剪枝做准备 + dfs(candidates, 0, target) return res } -func backtracking(startIndex,sum,target int,candidates,trcak []int,res *[][]int){ - //终止条件 - if sum==target{ - tmp:=make([]int,len(trcak)) - //拷贝 - copy(tmp,trcak) - //放入结果集 - *res=append(*res,tmp) + +func dfs(candidates []int, start int, target int) { + if target == 0 { // target 不断减小,如果为0说明达到了目标值 + tmp := make([]int, len(path)) + copy(tmp, path) + res = append(res, tmp) return } - //回溯 - for i:=startIndex;istartIndex&&candidates[i]==candidates[i-1]{ - continue + for i := start; i < len(candidates); i++ { + if candidates[i] > target { // 剪枝,提前返回 + break } - //更新路径集合和sum - trcak=append(trcak,candidates[i]) - sum+=candidates[i] - backtracking(i+1,sum,target,candidates,trcak,res) - //回溯 - trcak=trcak[:len(trcak)-1] - sum-=candidates[i] + // i != start 限制了这不对深度遍历到达的此值去重 + if i != start && candidates[i] == candidates[i-1] { // 去重 + continue + } + path = append(path, candidates[i]) + dfs(candidates, i+1, target - candidates[i]) + path = path[:len(path) - 1] } } ```