1760. 袋子里最少数目的球

1760. 袋子里最少数目的球

题解：
二分
1.题目可以转换为：操作 maxOperations 次，每个袋子的最大值小于等于 ans，求这个ans的最小值
2.可以从[1, nums数组的最大值]这个区间开始二分
3.判断函数：

• nums[i] <= ans opt = 0
• nums[i] > ans opt = (nums[i] - 1) / ans

class Solution {
    public int minimumSize(int[] nums, int maxOperations) {
        int left = 1;
        int right = 0;
        int n = nums.length;
        for (int i = 0; i < n; i++) {
            right = Math.max(right, nums[i]);
        }
        while (left < right) {
            int mid = (left  + right ) / 2 ;
            if (check(mid,nums,maxOperations)) left = mid + 1;
            else right = mid;
        }
        return left;
    }

    private boolean check(int mid, int[] nums, int maxOperations) {
        int opt = 0;
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] > mid) {
                opt += (nums[i] - 1) / mid;
            }
        }
        return opt > maxOperations;
    }
}