leetcode/0301.Remove-Invalid-Parentheses/README.md
Given a string s that contains parentheses and letters, remove the minimum number of invalid parentheses to make the input string valid.
Return all the possible results. You may return the answer in any order.
Example 1:
Input: s = "()())()"
Output: ["(())()","()()()"]
Example 2:
Input: s = "(a)())()"
Output: ["(a())()","(a)()()"]
Example 3:
Input: s = ")("
Output: [""]
Constraints:
给你一个由若干括号和字母组成的字符串 s ,删除最小数量的无效括号,使得输入的字符串有效。
返回所有可能的结果。答案可以按 任意顺序 返回。
说明:
回溯和剪枝
package leetcode
var (
res []string
mp map[string]int
n int
length int
maxScore int
str string
)
func removeInvalidParentheses(s string) []string {
lmoves, rmoves, lcnt, rcnt := 0, 0, 0, 0
for _, v := range s {
if v == '(' {
lmoves++
lcnt++
} else if v == ')' {
if lmoves != 0 {
lmoves--
} else {
rmoves++
}
rcnt++
}
}
n = len(s)
length = n - lmoves - rmoves
res = []string{}
mp = make(map[string]int)
maxScore = min(lcnt, rcnt)
str = s
backtrace(0, "", lmoves, rmoves, 0)
return res
}
func backtrace(i int, cur string, lmoves int, rmoves int, score int) {
if lmoves < 0 || rmoves < 0 || score < 0 || score > maxScore {
return
}
if lmoves == 0 && rmoves == 0 {
if len(cur) == length {
if _, ok := mp[cur]; !ok {
res = append(res, cur)
mp[cur] = 1
}
return
}
}
if i == n {
return
}
if str[i] == '(' {
backtrace(i+1, cur+string('('), lmoves, rmoves, score+1)
backtrace(i+1, cur, lmoves-1, rmoves, score)
} else if str[i] == ')' {
backtrace(i+1, cur+string(')'), lmoves, rmoves, score-1)
backtrace(i+1, cur, lmoves, rmoves-1, score)
} else {
backtrace(i+1, cur+string(str[i]), lmoves, rmoves, score)
}
}
func min(a, b int) int {
if a < b {
return a
}
return b
}