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update 0040.组合总和II: 修改错字,更换 go 代码
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@ -110,13 +110,13 @@ if (sum == target) {
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}
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```
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`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。
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`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。
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* **单层搜索的逻辑**
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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。
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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。
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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。
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**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]**。
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@ -438,76 +438,74 @@ class Solution:
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**使用used数组**
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```go
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var (
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res [][]int
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path []int
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used []bool
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)
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func combinationSum2(candidates []int, target int) [][]int {
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var trcak []int
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var res [][]int
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var history map[int]bool
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history=make(map[int]bool)
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sort.Ints(candidates)
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backtracking(0,0,target,candidates,trcak,&res,history)
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res, path = make([][]int, 0), make([]int, 0, len(candidates))
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used = make([]bool, len(candidates))
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sort.Ints(candidates) // 排序,为剪枝做准备
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dfs(candidates, 0, target)
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return res
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}
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func backtracking(startIndex,sum,target int,candidates,trcak []int,res *[][]int,history map[int]bool){
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//终止条件
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if sum==target{
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tmp:=make([]int,len(trcak))
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copy(tmp,trcak)//拷贝
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*res=append(*res,tmp)//放入结果集
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func dfs(candidates []int, start int, target int) {
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if target == 0 { // target 不断减小,如果为0说明达到了目标值
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tmp := make([]int, len(path))
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copy(tmp, path)
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res = append(res, tmp)
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return
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}
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if sum>target{return}
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//回溯
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// used[i - 1] == true,说明同一树枝candidates[i - 1]使用过
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// used[i - 1] == false,说明同一树层candidates[i - 1]使用过
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for i:=startIndex;i<len(candidates);i++{
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if i>0&&candidates[i]==candidates[i-1]&&history[i-1]==false{
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continue
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for i := start; i < len(candidates); i++ {
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if candidates[i] > target { // 剪枝,提前返回
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break
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}
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//更新路径集合和sum
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trcak=append(trcak,candidates[i])
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sum+=candidates[i]
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history[i]=true
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//递归
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backtracking(i+1,sum,target,candidates,trcak,res,history)
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//回溯
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trcak=trcak[:len(trcak)-1]
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sum-=candidates[i]
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history[i]=false
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// used[i - 1] == true,说明同一树枝candidates[i - 1]使用过
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// used[i - 1] == false,说明同一树层candidates[i - 1]使用过
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if i > 0 && candidates[i] == candidates[i-1] && used[i-1] == false {
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continue
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}
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path = append(path, candidates[i])
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used[i] = true
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dfs(candidates, i+1, target - candidates[i])
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used[i] = false
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path = path[:len(path) - 1]
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}
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}
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```
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**不使用used数组**
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```go
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var (
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res [][]int
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path []int
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)
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func combinationSum2(candidates []int, target int) [][]int {
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var trcak []int
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var res [][]int
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sort.Ints(candidates)
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backtracking(0,0,target,candidates,trcak,&res)
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res, path = make([][]int, 0), make([]int, 0, len(candidates))
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sort.Ints(candidates) // 排序,为剪枝做准备
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dfs(candidates, 0, target)
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return res
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}
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func backtracking(startIndex,sum,target int,candidates,trcak []int,res *[][]int){
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//终止条件
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if sum==target{
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tmp:=make([]int,len(trcak))
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//拷贝
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copy(tmp,trcak)
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//放入结果集
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*res=append(*res,tmp)
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func dfs(candidates []int, start int, target int) {
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if target == 0 { // target 不断减小,如果为0说明达到了目标值
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tmp := make([]int, len(path))
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copy(tmp, path)
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res = append(res, tmp)
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return
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}
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//回溯
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for i:=startIndex;i<len(candidates) && sum+candidates[i]<=target;i++{
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// 若当前树层有使用过相同的元素,则跳过
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if i>startIndex&&candidates[i]==candidates[i-1]{
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continue
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for i := start; i < len(candidates); i++ {
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if candidates[i] > target { // 剪枝,提前返回
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break
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}
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//更新路径集合和sum
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trcak=append(trcak,candidates[i])
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sum+=candidates[i]
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backtracking(i+1,sum,target,candidates,trcak,res)
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//回溯
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trcak=trcak[:len(trcak)-1]
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sum-=candidates[i]
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// i != start 限制了这不对深度遍历到达的此值去重
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if i != start && candidates[i] == candidates[i-1] { // 去重
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continue
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}
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path = append(path, candidates[i])
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dfs(candidates, i+1, target - candidates[i])
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path = path[:len(path) - 1]
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}
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}
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```
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