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update 0216.组合总和III: 替换 go 代码
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@ -36,7 +36,7 @@
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想到这一点了,做过[77. 组合](https://programmercarl.com/0077.组合.html)之后,本题是简单一些了。
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本题k相当于了树的深度,9(因为整个集合就是9个数)就是树的宽度。
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本题k相当于树的深度,9(因为整个集合就是9个数)就是树的宽度。
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例如 k = 2,n = 4的话,就是在集合[1,2,3,4,5,6,7,8,9]中求 k(个数) = 2, n(和) = 4的组合。
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@ -380,29 +380,32 @@ class Solution:
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回溯+减枝
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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 combinationSum3(k int, n int) [][]int {
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var track []int// 遍历路径
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var result [][]int// 存放结果集
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backTree(n,k,1,&track,&result)
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return result
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res, path = make([][]int, 0), make([]int, 0, k)
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dfs(k, n, 1, 0)
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return res
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}
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func backTree(n,k,startIndex int,track *[]int,result *[][]int){
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if len(*track)==k{
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var sum int
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tmp:=make([]int,k)
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for k,v:=range *track{
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sum+=v
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tmp[k]=v
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}
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if sum==n{
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*result=append(*result,tmp)
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func dfs(k, n int, start int, sum int) {
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if len(path) == k {
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if sum == n {
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tmp := make([]int, k)
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copy(tmp, path)
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res = append(res, tmp)
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}
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return
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}
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for i:=startIndex;i<=9-(k-len(*track))+1;i++{//减枝(k-len(*track)表示还剩多少个可填充的元素)
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*track=append(*track,i)//记录路径
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backTree(n,k,i+1,track,result)//递归
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*track=(*track)[:len(*track)-1]//回溯
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for i := start; i <= 9; i++ {
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if sum + i > n || 9-i+1 < k-len(path) {
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break
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}
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path = append(path, i)
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dfs(k, n, i+1, sum+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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