代码随想录算法训练营|day29
第七章 回溯算法
- 491.递增子序列
- 46.全排列
- 47.全排列II
- 代码随想录文章详解
- 总结
491.递增子序列
同层去重，只需保证当前层元素不重复即可【前仆后继的感觉】
func findSubsequences(nums []int) [][]int {
res, path := [][]int{}, []int{}
var help func(nums []int, startIndex int)
help = func(nums []int, startIndex int) {
if len(path) > 1 {
tmp := make([]int, len(path))
copy(tmp, path)
res = append(res, tmp)
}
used := make(map[int]bool, len(nums))
for i := startIndex; i < len(nums); i++ {
if used[nums[i]]|| len(path) > 0 && nums[i] < path[len(path)-1] {
continue
}
path = append(path, nums[i])
used[nums[i]] = true
help(nums, i+1)
path = path[:len(path)-1]
}
}
help(nums, 0)
return res
}
46.全排列
used数组，记录当前元素是否在path路径中存在，若存在则跳过。一个值在一条path路径中只能被选取一次
func permute(nums []int) [][]int {
res := [][]int{}
path := []int{}
used := make([]bool, len(nums))
var help func(nums []int, index int)
help = func(nums []int, index int) {
if len(path) == len(nums) {
tmp := make([]int, len(path))
copy(tmp, path)
res = append(res, tmp)
}
for i := 0; i < len(nums); i++ {
if used[i] == true {
continue
}
path = append(path, nums[i])
used[i] = true
help(nums, i+1)
path = path[:len(path)-1]
used[i] = false
}
}
help(nums, 0)
return res
}
47.全排列II
上题扩展，有重复元素。先排序，然后同层剪枝
同层剪枝：i > 0 && nums[i] == nums[i-1] && used[i-1] == false跳过，
当nums[i-1]作为某一层元素，其所有的递归选择已经穷尽，回溯时才会有used[i - 1]==false。因此，当nums[i] == nums[i-1]跳过，否则nums[i]递归得到的结果与nums[i - 1]结果重复
func permuteUnique(nums []int) [][]int {
res := [][]int{}
path := []int{}
used := make([]bool, len(nums))
sort.Ints(nums)
var help func(nums []int, index int)
help = func(nums []int, index int) {
if len(path) == len(nums) {
tmp := make([]int, len(path))
copy(tmp, path)
res = append(res, tmp)
return
}
for i := 0; i < len(nums); i++ {
if i > 0 && nums[i] == nums[i-1] && used[i-1] == false || used[i] == true {
continue
}
used[i] = true
path = append(path, nums[i])
help(nums, i+1)
path = path[:len(path)-1]
used[i] = false
}
}
help(nums, 0)
return res
}
代码随想录文章详解
491.递增子序列
46.全排列
47.全排列II
总结
常做常新是怎么回事
代码随想录图示帮助好大