func decodeString(s string) string {
if len(s) == 0 {
return ""
}
stack := make([]byte, 0)
for i := 0; i < len(s); i++ {
switch s[i] {
case ']':
temp := make([]byte, 0)
for len(stack) != 0 && stack[len(stack)-1] != '[' {
v := stack[len(stack)-1]
stack = stack[:len(stack)-1]
temp = append(temp, v)
}
// pop '['
stack = stack[:len(stack)-1]
// pop num
idx := 1
for len(stack) >= idx && stack[len(stack)-idx] >= '0' && stack[len(stack)-idx] <= '9' {
idx++
}
// 注意索引边界
num := stack[len(stack)-idx+1:]
stack = stack[:len(stack)-idx+1]
count, _ := strconv.Atoi(string(num))
for j := 0; j < count; j++ {
// 把字符正向放回到栈里面
for j := len(temp) - 1; j >= 0; j-- {
stack = append(stack, temp[j])
}
}
default:
stack = append(stack, s[i])
}
}
return string(stack)
}
字符串解码
通过此rep的模版无法通过，具体算法如下：