英文:
Combination Sum in Go
问题
给定一个数组:[1,2]和一个目标值:4
找到总和等于目标值的解集
在这种情况下:
[1,1,1,1]
[1,1,2]
[2,2]
这是Leetcode上的一个问题,我使用递归来解决它。
在第18行,当总和等于目标值时,我打印出每种情况。
输出结果为:
[1,1,1,1]
[1,1,2]
[2,2]
这就是我想要的答案!
但为什么最终的答案(二维数组)是:
[[1,1,1,2],[1,1,2],[2,2]]
期望的答案是:[[1,1,1,1],[1,1,2],[2,2]]
请帮我找出代码中的错误。谢谢你的时间。
英文:
/*
Given an array: [1,2] and a target: 4
Find the solution set that adds up to the target
in this case:
[1,1,1,1]
[1,1,2]
[2,2]
*/
import "sort"
func combinationSum(candidates []int, target int) [][]int {
sort.Ints(candidates)
return combine(0, target, []int{}, candidates)
}
func combine(sum int, target int, curComb []int, candidates []int) [][]int {
var tmp [][]int
var result [][]int
if sum == target {
fmt.Println(curComb)
return [][]int{curComb}
} else if sum < target {
for i,v := range candidates {
tmp = combine(sum+v, target, append(curComb, v), candidates[i:])
result = append(result,tmp...)
}
}
return result
}
This is a problem in Leetcode and I use recursion to solve it.
In line 18, I print every case when the sum is equal to the target.
The output is :
[1,1,1,1]
[1,1,2]
[2,2]
And that is the answer that I want!
But why is the final answer (two-dimensional):
[[1,1,1,2],[1,1,2],[2,2]]
Expected answer is : [[1,1,1,1],[1,1,2],[2,2]]
Please help me find the mistake in the code. Thanks for your time.
答案1
得分: 2
这是因为切片的工作方式导致的。切片对象是对底层数组的引用,同时包含切片的长度、切片在数组中的起始位置的指针以及切片的容量。切片的容量是从切片开始到数组末尾的元素数量。当你向切片追加元素时,如果有足够的容量可以存放新元素,它会被添加到现有的数组中。然而,如果容量不足,append
函数会分配一个新的数组并将元素复制过去。新数组会分配额外的容量,以避免每次追加都需要重新分配内存。
在你的 for
循环中,当 curComb
是 [1, 1, 1]
时,它的容量是 4。在循环的连续迭代中,你追加了 1 和 2,由于数组中有足够的空间来容纳新元素,所以不会触发重新分配内存。当 curComb
变为 [1, 1, 1, 1]
时,它被放入结果列表中,但在下一次 for
循环的迭代中,append
函数会修改最后一个元素为 2(记住它们共享同一个底层数组),所以在最后打印结果时你看到的是 [1, 1, 1, 2]
。
解决这个问题的方法是在和等于目标值时返回 curComb
的副本:
if sum == target {
fmt.Println(curComb)
tmpCurComb := make([]int, len(curComb))
copy(tmpCurComb, curComb)
return [][]int{tmpCurComb}
}
这篇文章 对切片的工作原理进行了很好的解释。
英文:
This happens because of the way slices work. A slice object is a reference to an underlying array, along with the length of the slice, a pointer to the start of the slice in the array, and the slice's capacity. The capacity of a slice is the number of elements from the beginning of the slice to the end of the array. When you append to a slice, if there is available capacity for the new element, it is added to the existing array. However, if there isn't sufficient capacity, append
allocates a new array and copies the elements. The new array is allocated with extra capacity so that an allocation isn't required for every append.
In your for
loop, when curComb
is [1, 1, 1]
, its capacity is 4. On successive iterations of the loop, you append 1 and then 2, neither of which causes a reallocation because there's enough room in the array for the new element. When curComb
is [1, 1, 1, 1]
, it is put on the results list, but in the next iteration of the for
loop, the append
changes the last element to 2 (remember that it's the same underlying array), so that's what you see when you print the results at the end.
The solution to this is to return a copy of curComb
when the sum equals the target:
if sum == target {
fmt.Println(curComb)
tmpCurComb := make([]int, len(curComb))
copy(tmpCurComb, curComb)
return [][]int{tmpCurComb}
This article gives a good explanation of how slices work.
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