LeetCode1334 value within a minimum distance of neighboring cities

OJ

The meaning of problems
given the difficulty of the work of a group of sequence, a given number of days. The completion of the i-th job, must first complete (0, i-1) work, difficulty of the work day is the day the work done in the most difficult that a daily difficulty of the group and for the completion of this work difficult the sum must have to work every day.

Solution to a problem
typical of dynamic programming problem, with dp [i] [j] to represent the minimum difficulty i j days to complete a total of two tasks. How recursive relationship? i j day to complete the task, as long as this j tasks assigned to it these days, the value of minimum allocation, assuming k i days to complete the task, then the first i-1 jk days necessary to complete a task, k> = i, k <= the number of tasks
such days to complete j i can be calculated minimum task difficulty of

public int minDifficulty(int[] jobDifficulty, int d) {
	// 工作数量必须大于天数
    if (jobDifficulty.length < d) {
        return -1;
    }
    int[][] dp = new int[d + 1][jobDifficulty.length + 1];
    dp[1][1] = jobDifficulty[0];
    // 初始化第一天完成i个工作的最小难度
    for (int i = 2; i <= jobDifficulty.length; i++) {
        dp[1][i] = Math.max(dp[1][i - 1], jobDifficulty[i - 1]);
    }
    // map用来记录工作数组中i到j这个区间的最大值
    int[][] map = new int[jobDifficulty.length][jobDifficulty.length];
    for (int i = 0; i < jobDifficulty.length; i++) {
        map[i][i] = jobDifficulty[i];
        for (int j = i + 1; j < jobDifficulty.length; j++) {
            map[i][j] = Math.max(jobDifficulty[j], map[i][j - 1]);
        }
    }
    // 外层循环天数
    for (int i = 2; i <= d; i++) {
   		// 内层循环完成的任务数，第i天要完成的任务数必须大于等于i
        for (int j = i; j <= jobDifficulty.length; j++) {
            dp[i][j] = Integer.MAX_VALUE;
            // 总共要完成j个任务，i-1天分配部分任务，i天分配部分任务
            // 计算如何分配才能使难度最小
            for (int k = i - 1; k < j; k++) {
                dp[i][j] = Math.min(dp[i - 1][k] + map[k][j - 1], dp[i][j]);
            }
        }
    }
    // 返回d天完成指定任务的最小难度
    return dp[d][jobDifficulty.length];
}