示例 1：
输入：bloomDay = [1,10,3,10,2], m = 3, k = 1输出：3解释：让我们一起观察这三天的花开过程，x 表示花开，而 _ 表示花还未开。现在需要制作 3 束花，每束只需要 1 朵。1 天后：[x, _, _, _, _]   // 只能制作 1 束花2 天后：[x, _, _, _, x]   // 只能制作 2 束花3 天后：[x, _, x, _, x]   // 可以制作 3 束花，答案为 3
来源：力扣（LeetCode）链接：https://leetcode-cn.com/problems/minimum-number-of-days-to-make-m-bouquets著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

public class DayCode {    public static void main(String[] args) {        int[] bloomDay = new int[]{1, 10, 3, 10, 2};        int m = 3;        int k = 1;        int ans = new DayCode().minDays(bloomDay, m, k);        System.out.println(ans);    }
    public int minDays(int[] bloomDay, int m, int k) {        int n = bloomDay.length;        if (n < m * k) {            return -1;        }        int low = Integer.MAX_VALUE;        int high = 0;        for (int i = 0; i < n; i++) {            low = Math.min(low, bloomDay[i]);            high = Math.max(high, bloomDay[i]);        }        while (low < high) {            int days = low + (high - low) / 2;            if (help(bloomDay, days, m, k)) {                high = days;            } else {                low = days + 1;            }        }
        return low;    }
    private boolean help(int[] bloomDay, int days, int m, int k) {        int flowers = 0;        int bouquets = 0;        int n = bloomDay.length;        for (int i = 0; i < n && bouquets < m; i++) {            if (bloomDay[i] <= days) {                flowers++;                if (flowers == k) {                    bouquets++;                    flowers = 0;                }            } else {                flowers = 0;            }        }
        return bouquets >= m;    }}