[LeetCode] 1717. Maximum Score From Removing Substrings

You are given a string s and two integers x and y. You can perform two types of operations any number of times.

• Remove substring "ab" and gain x points.
  • For example, when removing "ab" from "cabxbae" it becomes "cxbae".
• Remove substring "ba" and gain y points.
  • For example, when removing "ba" from "cabxbae" it becomes "cabxe".

Return the maximum points you can gain after applying the above operations on s.

Example 1:

Input: s = "cdbcbbaaabab", x = 4, y = 5
Output: 19
Explanation:
- Remove the "ba" underlined in "cdbcbbaaabab". Now, s = "cdbcbbaaab" and 5 points are added to the score.
- Remove the "ab" underlined in "cdbcbbaaab". Now, s = "cdbcbbaa" and 4 points are added to the score.
- Remove the "ba" underlined in "cdbcbbaa". Now, s = "cdbcba" and 5 points are added to the score.
- Remove the "ba" underlined in "cdbcba". Now, s = "cdbc" and 5 points are added to the score.
Total score = 5 + 4 + 5 + 5 = 19.

Example 2:

Input: s = "aabbaaxybbaabb", x = 5, y = 4
Output: 20

Constraints:

• 1 <= s.length <= 105
• 1 <= x, y <= 104
• s consists of lowercase English letters.

删除子字符串的最大得分。

给你一个字符串 s 和两个整数 x 和 y 。你可以执行下面两种操作任意次。

删除子字符串 "ab" 并得到 x 分。
比方说，从 "cabxbae" 删除 ab ，得到 "cxbae" 。
删除子字符串"ba" 并得到 y 分。
比方说，从 "cabxbae" 删除 ba ，得到 "cabxe" 。
请返回对 s 字符串执行上面操作若干次能得到的最大得分。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/maximum-score-from-removing-substrings
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这道题总体的思路是贪心，这里我给出一个需要用到额外空间的做法。我需要用到两个stack。首先我预处理一下input，这样我知道到底是ab组合得分更高还是ba组合得分更高。接着我开始遍历input字符串，我优先处理得分高的组合，比如如果得分高的组合是ab的话，当我遇到ab这样的子串，我优先处理，然后把它的得分加到res中；其他不是ab的字母全都加到第一个栈中，记为stack1。

接着我再从stack1中把所有字符串倒出来，因为栈是先进后出的，所以如果一开始我处理的是ab的话，那么我现在正好可以处理ba这种子串了。从stack1中弹出的字符我放到另一个栈，记为stack2。这样我可以利用stack2获取到ba这种组合同时记录它的得分。

时间O(n)

空间O(n)

Java实现

 1 class Solution {
 2     public int maximumGain(String s, int x, int y) {
 3         Stack<Character> stack1 = new Stack<>();
 4         Stack<Character> stack2 = new Stack<>();
 5         int res = 0;
 6         int max = Math.max(x, y);
 7         int min = Math.min(x, y);
 8         char first = (x > y) ? 'a' : 'b';
 9         char second = (x > y) ? 'b' : 'a';
10         for (char c : s.toCharArray()) {
11             if (!stack1.isEmpty() && stack1.peek() == first && c == second) {
12                 stack1.pop();
13                 res += max;
14             } else {
15                 stack1.push(c);
16             }
17         }
18 
19         while (!stack1.isEmpty()) {
20             char c = stack1.pop();
21             if (!stack2.isEmpty() && stack2.peek() == first && c == second) {
22                 stack2.pop();
23                 res += min;
24             } else {
25                 stack2.push(c);
26             }
27         }
28         return res;
29     }
30 }

LeetCode 题目总结

标签：得分,ab,1717,ba,Removing,Maximum,points,stack2,stack1
来源： https://www.cnblogs.com/cnoodle/p/14477998.html