LeetCode每日一题(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.

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假如 x < y, 那我们就优先消耗’ba’， 如果 x > y 则我们优先消耗’ab’。 其实我们需要考虑的就以下四中情况， ‘abab’, ‘baba’, ‘bbaa’, ‘aabb’

---

impl Solution {
    
    pub fn maximum_gain(s: String, x: i32, y: i32) -> i32 {
    
        let mut stack: Vec<char> = Vec::new();
        let mut ans = 0;
        for c in s.chars() {
    
            if stack.is_empty() {
    
                stack.push(c);
                continue;
            }
            if c == if x <= y {
     'a' } else {
     'b' } {
    
                let last = stack.pop().unwrap();
                if last == if x <= y {
     'b' } else {
     'a' } {
    
                    ans += if x <= y {
     y } else {
     x };
                    continue;
                }
                stack.push(last);
                stack.push(c);
                continue;
            }
            stack.push(c);
        }
        let mut stack2 = Vec::new();
        for c in stack {
    
            if stack2.is_empty() {
    
                stack2.push(c);
                continue;
            }
            let last = stack2.pop().unwrap();
            if c == 'a' && last == 'b' {
    
                ans += y;
                continue;
            }
            if c == 'b' && last == 'a' {
    
                ans += x;
                continue;
            }
            stack2.push(last);
            stack2.push(c);
        }
        ans
    }
}