算法专题：栈-CFANZ编程社区

class Solution {
    public String removeDuplicates(String s) {
        // 字符串模拟栈结构
        StringBuilder stack = new StringBuilder();
        for(int i = 0; i < s.length(); i++) {
            char ch = s.charAt(i);
            int len = stack.length();
            // 栈为空 || ch 和栈顶元素不相同 => 入栈
            if(len == 0 || stack.charAt(len - 1) != ch) stack.append(ch);
            // ch 和栈顶元素相同 => 出栈
            else stack.deleteCharAt(len - 1);
        }
        return stack.toString();
    }
}

2. 844. 比较含退格的字符串

2.1 算法原理

本题思路与第一题相同,

当 s[i] 为 '#' 时, 出栈
当 s[i] 不为 '#' 时, 入栈
最后比较经过退格处理后的两个字符串是否相同

2.2 算法原理

class Solution {
    public boolean backspaceCompare(String s, String t) {
        return strOpera(s).equals(strOpera(t));
    }
    public String strOpera(String s) {
        StringBuilder stringBuilder = new StringBuilder();
        for (int i = 0; i < s.length(); i++) {
            char ch = s.charAt(i);
            if (ch == '#') {
                // 出栈
                if(stringBuilder.length() >= 1) stringBuilder.deleteCharAt(stringBuilder.length() - 1);
            }
            else {
                // 入栈
                stringBuilder.append(ch);
            }
        }
        return stringBuilder.toString();
    }
}

3. 基本计算器 II

3.1 算法原理

解法: 利用栈模拟运算过程

分类讨论:

1. s[i] 是运算符 => op = s[i] (记录运算符)

2. s[i] 是数字字符 , 把数字提取出来: tmp += s[i] - '0' (可能s[i + 1]仍为数字字符, 需循环处理)

op == '+' => stack.push(tmp);
op == '-' => stack.push(-tmp);
op == '*' => stack.push(stack.pop() * tmp);
op == '/' => stack.push(stack.pop() / tmp);

3. 最终, 返回栈中所有元素的和

注意:
需要处理空格字符!!

3.2 算法代码

class Solution {
    public int calculate(String ss) {
        char[] s = ss.toCharArray();
        int n = s.length;
        char op = '+';
        int tmp = 0;
        Stack<Integer> stack = new Stack<>();
        int i = 0;
        while(i < n) {
            if(s[i] == ' ') {
                i++;
            }else if(isOperator(s[i])) {
                op = s[i++];
            }else {
                tmp = s[i] - '0';
                while(i + 1 < n && !isOperator(s[i + 1]) && s[i + 1] != ' ') {
                    tmp *= 10;
                    tmp += s[i + 1] - '0';
                    i++;
                }
                if(op == '+') stack.push(tmp);
                else if(op == '-') stack.push(-tmp);
                else if(op == '*') stack.push(stack.pop() * tmp);
                else stack.push(stack.pop() / tmp);
                i++;
            }
        }
        int ret = 0;
        for(int x : stack)  ret += x;
        return ret;
    }
    public boolean isOperator(char ch) {
        if(ch == '+' || ch == '-' || ch == '*' ||ch == '/') return true;
        return false;
    }
}

4. 字符串解码

4.1 算法原理

解法: 栈模拟

遇见数字 => 放入"数字栈"
遇见 '[' => 将 '[' 后的字母字符串提取出来, 放入 "字符串栈"
遇见 ']' => 取 "字符串栈" 栈顶元素和 "数字栈" 栈顶元素进行解析,并拼接到 "字符串栈" 栈顶元素后
遇见单独的字符串 => 拼接到 "字符串栈" 栈顶元素后

4.2 算法代码

class Solution {
    public String decodeString(String ss) {
        char[] s = ss.toCharArray();
        int n = s.length;
        Stack<Integer> intStack = new Stack<>();
        Stack<String> stringStack = new Stack<>();
        // 预处理 => 当栈为空时
        stringStack.push("");
        int i = 0;
        while(i < n) {
            if(s[i] > '0' && s[i] < '9') {
                int tmp = 0;
                while(i < n && s[i] >= '0' && s[i] <= '9') {
                    tmp = tmp * 10 + (s[i++] - '0');
                }
                intStack.push(tmp);
            }else if(s[i] == '[') {
                i++;
                StringBuilder sb = new StringBuilder();
                while(i < n && s[i] >= 'a' && s[i] <= 'z') sb.append(s[i++]);
                stringStack.push(sb.toString());
            }else if(s[i] == ']') {
                StringBuilder sb = new StringBuilder();
                int x = intStack.pop();
                String str = stringStack.pop();
                while(x-- != 0) {
                    sb.append(str);
                }
                stringStack.push(stringStack.pop() + sb.toString());
                i++;
            }else {
                StringBuilder sb = new StringBuilder();
                while(i < n && s[i] >= 'a' && s[i] <= 'z') sb.append(s[i++]);
                stringStack.push(stringStack.pop() + sb.toString());               
            }
        }
        return stringStack.pop();
    }
}

5. 验证栈序列

5.1 算法原理

解法: 使用栈模拟.

一边进栈, 一边判断是否需要出栈.

对 pushed 中的元素进行进栈操作, 将栈顶元素和 popped[i] 比较, 若不相等则一直进栈, 若相等则出栈, 并判断是否需要循环出栈.

5.2 算法代码

class Solution {
    public boolean validateStackSequences(int[] pushed, int[] popped) {
        Stack<Integer> stack = new Stack<>();
        int i = 0, n = popped.length;
        for(int x : pushed) {
            stack.push(x);
            while(!stack.isEmpty() && stack.peek() == popped[i]) {
                stack.pop();
                i++;
            }
        }
        return i == n;
    }
}

END