LeetCode 394. 字符串解码
题目描述
思路分析
栈
参考代码
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func decodeString(s string) string {
numStack := make([]int, 0)
strStack := make([]string, 0)
curStr, num := "", 0
for i := 0; i < len(s); i++ {
c := s[i]
if c >= '0' && c <= '9' {
num = num*10 + int(c-'0')
} else if c == '[' {
numStack = append(numStack, num)
strStack = append(strStack, curStr)
curStr, num = "", 0
} else if c == ']' {
count := numStack[len(numStack)-1]
numStack = numStack[:len(numStack)-1]
preStr := strStack[len(strStack)-1]
strStack = strStack[:len(strStack)-1]
curStr = preStr + strings.Repeat(curStr, count)
} else {
curStr += string(c)
}
}
return curStr
}
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func decodeString(s string) string {
var stack []string
curNum := 0
curStr := ""
for _, cur := range s {
if cur >= '0' && cur <= '9' {
curNum = curNum*10 + int(cur-'0') // 构建数字
} else if cur == '[' {
// 将当前数字和字符串状态压入栈中
stack = append(stack, curStr)
stack = append(stack, strconv.Itoa(curNum))
curStr = ""
curNum = 0
} else if cur == ']' {
// 弹出栈顶的字符串和数字
numStr := stack[len(stack)-1]
stack = stack[:len(stack)-1]
preStr := stack[len(stack)-1]
stack = stack[:len(stack)-1]
// 重复构建字符串
repeatNum, _ := strconv.Atoi(numStr)
for i := 0; i < repeatNum; i++ {
preStr += curStr
}
curStr = preStr
} else {
// 处理普通字符
curStr += string(cur)
}
}
return curStr
}
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