（算法题）Day-20230308

Excel 表列序号

从后往前 乘26

class Solution {
    public int titleToNumber(String columnTitle) {
        int len = columnTitle.length();
        int num = 0;
        int res=0;
        for(int i=len-1;i>=0;i--){
            int nofc = columnTitle.charAt(i)-'A'+1;
            res += nofc*Math.pow(26,num);
            num+=1;
        }
        return res;
    }
}

Excel表列名称

细节很多！！！是A、B、C开始，相当于从1、2、3开始没有0，所以每次除 都得先减一！

一直除26，每次将余数 转为字符，注意最后需要进行翻转

class Solution {
    public String convertToTitle(int columnNumber) {
        StringBuilder sb = new StringBuilder();
        while(columnNumber > 0){
            columnNumber--;
            sb.append((char)(columnNumber%26 + 'A'));
            columnNumber/=26;
        }
        sb.reverse();	
        return sb.toString();
    }
}

验证栈序列

模拟 入栈和出栈

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

字典序排数

递归法

class Solution {
    List<Integer> ans = new ArrayList<>();
    public List<Integer> lexicalOrder(int n) {
        for (int i = 1; i <= 9; i++) dfs(i, n);
        return ans;
    }
    void dfs(int cur, int limit) {
        if (cur > limit) return ;
        ans.add(cur);
        for (int i = 0; i <= 9; i++) dfs(cur * 10 + i, limit);
    }
}

迭代法（代码非常优雅）

class Solution {
    public List<Integer> lexicalOrder(int n) {
        List<Integer> ans = new ArrayList<>();
        for (int i = 0, j = 1; i < n; i++) {
            ans.add(j);
            if (j * 10 <= n) {
                j *= 10;
            } else {
                while (j % 10 == 9 || j + 1 > n) j /= 10;
                j++;
            }
        }
        return ans;
    }
}