Java 第 39 课 316. 去除重复字母 321. 拼接最大数

• 316. 去除重复字母
• 321. 拼接最大数

Leetcode

class Solution:
    def removeDuplicateLetters(self, s: str) -> str:
        q, vis, count = [], set(), Counter(s)
        for ch in s:
            if ch not in vis:
                while q and q[-1] > ch and count[q[-1]]:
                    # q.pop() # 前一个不唯一，并且比较大，删除。
                    vis.remove(q.pop())            
                q.append(ch)
                vis.add(ch)
            count[ch] -= 1
        return ''.join(q)

class Solution {
    public String removeDuplicateLetters(String s) {
        boolean[] vis = new boolean[26];
        int[] num = new int[26];
        char[] arr = s.toCharArray();        
        for (char ch : arr){num[ch - 'a']++;}
        StringBuilder sb = new StringBuilder();
        for (char ch : arr){
            if (!vis[ch - 'a']){
                int i = sb.length() - 1;
                while (i > -1 && sb.charAt(i) > ch && num[sb.charAt(i) - 'a'] > 0){                               
                    vis[sb.charAt(i) - 'a'] = false;
                    sb.deleteCharAt(i);                     
                    i--;        
                }
                vis[ch - 'a'] = true;
                sb.append(ch);
            }
            num[ch - 'a'] -= 1;
        }
        return sb.toString();
    }
}

Leetcode

class Solution:
    def maxNumber(self, nums1, nums2, k):

        def pick(nums, k):
            q, drop = [], len(nums) - k
            for num in nums:
                while drop > 0 and q and q[-1] < num:
                    q.pop()
                    drop -= 1
                q.append(num)
            return q[:k]

        def merge(A, B):
            ans = []
            while A or B:
                bigger = A if A > B else B
                ans.append(bigger[0])
                bigger.pop(0)
            return ans

        return max(merge(pick(nums1, i), pick(nums2, k - i)) for i in range(k+1) if i <= len(nums1) and k-i <= len(nums2))