2511.最多可以摧毁的敌人城堡数目

链接：2511.最多可以摧毁的敌人城堡数目
难度：Easy
标签：数组、双指针
简介：请你返回 最多 可以摧毁的敌人城堡数目。

题解 1 - python

• 编辑时间：2023-09-02
• 执行用时：60ms
• 内存消耗：15.82MB
• 编程语言：python
• 解法介绍：同上。

class Solution:
    def captureForts(self, forts: List[int]) -> int:
        res = 0
        p0 = p1 = -1
        for i in range(len(forts)):
            fort = forts[i]
            if fort == 1:
                if p0 != -1 and p0 > p1:
                    res = max(res, i - p0 - 1)
                p1 = i
            elif fort == -1:
                if p1 != -1 and p1 > p0:
                    res = max(res, i - p1 - 1)
                p0 = i
        return res

题解 2 - cpp

• 编辑时间：2023-09-02
• 执行用时：4ms
• 内存消耗：7.41MB
• 编程语言：cpp
• 解法介绍：遍历。

class Solution {
public:
    int captureForts(vector<int>& forts) {
        int res = 0, p0 = -1, p1 = -1;
        for (int i = 0; i < forts.size(); i++) {
            int fort = forts[i];
            if (fort == 1) {
                if (p0 != -1 && p0 > p1) res = max(res, i - 1 - p0);
                p1 = i;
            } else if (fort == -1) {
                if (p1 != -1 && p1 > p0) res = max(res, i - 1 - p1);
                p0 = i;
            }
        }
        return res;
    }
};

题解 3 - rust

• 编辑时间：2023-09-02
• 内存消耗：1.93MB
• 编程语言：rust
• 解法介绍：同上。

impl Solution {
    pub fn capture_forts(forts: Vec<i32>) -> i32 {
        let mut res = 0i32;
        let (mut p0, mut p1) = (-1i32, -1i32);
        for i in 0..forts.len() {
            let fort = forts[i];
            if fort == 1 {
                if p0 != -1 && p0 > p1 {
                    res = res.max((i as i32) - 1 - p0);
                }
                p1 = i as i32;
            } else if fort == -1 {
                if p1 != -1 && p1 > p0 {
                    res = res.max((i as i32) - 1 - p1);
                }
                p0 = i as i32;
            }
        }
        res as i32
    }
}