bug修复 双指针 跳出循环中的循环
https://leetcode.cn/problems/backspace-string-compare/
844. Backspace String Compare
Given two strings s and t, return true if they are equal when both are typed into empty text editors. '#' means a backspace character.
Note that after backspacing an empty text, the text will continue empty.
Example 1:
Input: s = "ab#c", t = "ad#c"
Output: true
Explanation: Both s and t become "ac".
Example 2:
Input: s = "ab##", t = "c#d#"
Output: true
Explanation: Both s and t become "".
Example 3:
Input: s = "a#c", t = "b"
Output: false
Explanation: s becomes "c" while t becomes "b".
Constraints:
1 <= s.length, t.length <= 200
s and t only contain lowercase letters and '#' characters.
Follow up: Can you solve it in O(n) time and O(1) space?
func backspaceCompare(s string, t string) bool {
m, n := len(s), len(t)
p, q := m-1, n-1
for p > -1 && q > -1 {
i := 0
for p-i > -1 && s[p-i] == '#' {
i++
}
j := 0
for q-j > -1 && t[q-j] == '#' {
j++
}
p -= 2 * i
q -= 2 * j
if p < 0 && q < 0 {
return true
}
if p > -1 && q > -1 && s[p] != t[q] {
return false
}
p--
q--
}
return true
}
import "fmt"
func backspaceCompare(s string, t string) bool {
m, n := len(s), len(t)
p, q := m-1, n-1
for p > -1 && q > -1 {
for {
i := 0
for p-i > -1 && s[p-i] == '#' {
i++
}
j := 0
for q-j > -1 && t[q-j] == '#' {
j++
}
// offset #
u := 0
for ii := 0; ii < i; ii++ {
if p-i-ii > -1 && s[p-i-ii] == '#' {
u++
}
}
v := 0
for jj := 0; jj < j; jj++ {
if q-j-jj > 0 && t[q-j-jj] == '#' {
v++
}
}
if i+j == 0 {
break
}
p -= 2*i + 2*u // TODO 检查u经过的元素 循环中的循环
q -= 2*j + 2*v
}
if p > -1 && q > -1 {
if s[p] == t[q] {
p--
q--
continue
} else {
fmt.Println("2--")
return false
}
}
if p == -1 && q == -1 {
return true
}
if (p > -1 && s[p] != '#' && q < 0) || (q > -1 && t[q] != '#' && p < 0) {
fmt.Println("1--", p, " ", q)
return false
}
}
return true
}
循环中的循环
跳出
func backspaceCompare(s string, t string) bool {
skipS, skipT := 0, 0
i, j := len(s)-1, len(t)-1
for i >= 0 || j >= 0 {
for i >= 0 {
if s[i] == '#' {
skipS++
i--
} else if skipS > 0 {
skipS--
i--
} else {
break
}
}
for j >= 0 {
if t[j] == '#' {
skipT++
j--
} else if skipT > 0 {
skipT--
j--
} else {
break
}
}
if i >= 0 && j >= 0 {
if s[i] != t[j] {
return false
} else {
i--
j--
continue
}
} else if i >= 0 || j >= 0 {
return false
} else {
return true
}
}
return true
}
