贪心+前缀和解决连续数组问题

class Solution {
    public int findMaxLength(int[] nums) {
        int len = nums.length;
        if(len <= 1){
            return 0;
        }
        //前缀和
        int count = 0;
        //进行更新的
        int maxLen = 0;
        //哈希表记录前缀和对应的索引
        HashMap<Integer, Integer> index = new HashMap<>();
        //在索引为0之前默认为0
        index.put(0, -1);

        for(int i = 0; i < len; i++){
            if(nums[i] == 1) count++;
            if(nums[i] == 0) count--;

            //如果包含数值为0的前缀和，说明可以连续在一起，maxLen进行比较更新
            if(index.containsKey(count)){   
                maxLen = Math.max(maxLen, i - index.get(count));
            }else{
                //放入哈希表中
                index.put(count, i);
            }
        }
        return maxLen;
    }
}

美团笔试

import java.io.*;
public class Main{
    public static void main(String[] args) throws IOException{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(br.readLine().trim());
        char[] ef = br.readLine().trim().toCharArray();
        int max = 0, sum = 0;
        for(int i = 0; i < n; i++){
            if(ef[i] == 'E') sum++;
            if(ef[i] == 'F') sum--;
            max = Math.max(max, sum);
            sum = Math.max(sum, 0);
        }
        System.out.println(max);
    }
}