2511 Maximum Enemy Forts That Can Be Captured

You are given a 0-indexed integer array forts of length n representing the positions of several forts. forts[i] can be -1, 0, or 1 where:

• -1 represents there is no fort at the ith position.

• 0 indicates there is an enemy fort at the ith position.

• 1 indicates the fort at the ith the position is under your command.

Now you have decided to move your army from one of your forts at position i to an empty position j such that:

• 0 <= i, j <= n - 1

• The army travels over enemy forts only. Formally, for all k where min(i,j) < k < max(i,j), forts[k] == 0.

While moving the army, all the enemy forts that come in the way are captured.

Return the maximum number of enemy forts that can be captured. In case it is impossible to move your army, or you do not have any fort under your command, return 0.

Example 1:

Input: forts = [1,0,0,-1,0,0,0,0,1]
Output: 4
Explanation:
- Moving the army from position 0 to position 3 captures 2 enemy forts, at 1 and 2.
- Moving the army from position 8 to position 3 captures 4 enemy forts.
Since 4 is the maximum number of enemy forts that can be captured, we return 4.

Example 2:

Input: forts = [0,0,1,-1]
Output: 0
Explanation: Since no enemy fort can be captured, 0 is returned.

Constraints:

• 1 <= forts.length <= 1000

• -1 <= forts[i] <= 1

这题，要找1和-1或者-1和1之间有多少个0.用同向双指针，先用right每次往右扫一扫。找到一个非0的，就用last来记录一下。然后，如果right又指向非0的数字，判断一下是否跟last是一样的，一样的话，不处理，更新last位置。一样的情况表示我们是1，0，1 or -1，0，0，-1这些情况，所以跳过。最后，如果两个指向的是不同的数字，1，0，0，-1 or 1， 0， -1之类的情况，我们可以算个max。记住代入下标算。T: O(n), S:O(1)

public int captureForts(int[] forts) {
    if (forts == null || forts.length == 0) {
        return 0;
    }

    int max = 0;
    int last = -1;
    int right = 0;
    while (right < forts.length) {
        if (forts[right] == 0) {
            right++;
            continue;
        }

        if (last == -1) {
            last = right;
            right++;  
            continue;
        }

        if (forts[last] == forts[right]) {
            last = right;
            right++;  
            continue;
        } else {                
            max = Math.max(right - last - 1, max);
            last = right;
            right++;  
        }                      
    }
    return max;
}