LeetCode 处理用时最长的那个任务的员工

 解题思路：把第一个一维数组的两个元素分别定义为最大值和id，之后遍历进行判断

class Solution {
    public int hardestWorker(int n, int[][] logs) {
        int max = logs[0][1];
        int id = logs[0][0];
        for (int i = 1; i < logs.length; i++) {
            if(logs[i][1] - logs[i-1][1] > max){
                max = logs[i][1] - logs[i-1][1];
                id = logs[i][0];
            }else if(max == logs[i][1] - logs[i-1][1]){
                if(logs[i][0] > id){
                    id = id;
                }else{
                    id = logs[i][0];
                }
            }
        }
        return id;
    }
}