给定一个字符串 s ,请将 s 分割成一些子串,使每个子串都是 回文串 ,返回 s 所有可能的分割方案。
回文串 是正着读和反着读都一样的字符串。
示例 1:输入:s = "google" 输出:[["g","o","o","g","l","e"],["g","oo","g","l","e"],["goog","l","e"]]
示例 2:输入:s = "aab" 输出:[["a","a","b"],["aa","b"]]
示例 3:输入:s = "a" 输出:[["a"]
提示:1 <= s.length <= 16
s 仅由小写英文字母组成
注意:本题与主站 131 题相同:
1、回溯;时间复杂度O(n*2^n),空间复杂度O(n*2^n)
var res [][]string
func partition(s string) [][]string {
res = make([][]string, 0)
arr := make([]string, 0)
dfs(s, 0, arr)
return res
}
func dfs(s string, level int, arr []string) {
if level == len(s) {
temp := make([]string, len(arr))
copy(temp, arr)
res = append(res, temp)
return
}
for i := level; i < len(s); i++ {
str := s[level : i+1]
if judge(str) == true {
dfs(s, i+1, append(arr, str))
}
}
}
func judge(s string) bool {
for i := 0; i < len(s)/2; i++ {
if s[i] != s[len(s)-1-i] {
return false
}
}
return true
}
2、动态规划+回溯;时间复杂度O(n*2^n),空间复杂度O(n*2^n)
var res [][]string
var dp [][]bool
func partition(s string) [][]string {
res = make([][]string, 0)
arr := make([]string, 0)
dp = make([][]bool, len(s))
for r := 0; r < len(s); r++ {
dp[r] = make([]bool, len(s))
dp[r][r] = true
for l := 0; l < r; l++ {
if s[l] == s[r] && (r-l <= 2 || dp[l+1][r-1] == true) {
dp[l][r] = true
} else {
dp[l][r] = false
}
}
}
dfs(s, 0, arr)
return res
}
func dfs(s string, level int, arr []string) {
if level == len(s) {
temp := make([]string, len(arr))
copy(temp, arr)
res = append(res, temp)
return
}
for i := level; i < len(s); i++ {
str := s[level : i+1]
if dp[level][i] == true {
dfs(s, i+1, append(arr, str))
}
}
}
Medium题目,题目同leetcode 131.分割回文串
页面更新:2024-03-16
本站资料均由网友自行发布提供,仅用于学习交流。如有版权问题,请与我联系,QQ:4156828
© CopyRight 2020-2024 All Rights Reserved. Powered By 71396.com 闽ICP备11008920号-4
闽公网安备35020302034903号