分析：在整個數域範圍上二分查詢最小磁力的大小

　　　　　 然後記錄這個最小磁力可以在position上放置幾個球，根據球的數量與m的大小來決定二分範圍

class Solution {
    public int maxDistance(int[] position, int m) {
        int n = position.length;
        Arrays.sort(position);
        int l = 0, r = position[n-1];
        while(l <= r) { // 整個[l,r]範圍上查詢
            int
 
 mid = l + (r - l) / 2;
            int t = 1, pre = 0;
            for(int i = 1; i < n; i++) {
                if(position[i] - position[pre] < mid) continue;
                t++;
                pre = i;
            }
            if(t >= m) l = mid + 1; // 放多了或者剛好還可能有剩餘 距離加大
            else
 
 r = mid - 1; // 放少了 距離減小
        }
        return r;
    }
}

class Solution {
    public int minDays(int[] bloomDay, int m, int k) {
        int n = bloomDay.length;
        if(n < m * k) return -1;
        int l = 1, r = 1000000000;
        while(l <= r) {
            int mid = (l + r) >> 1;
            
 
int constant = 0, sum = 0;
            for(int num : bloomDay) {
                if(num <= mid) {
                    constant++;
                } else {
                    constant = 0;
                }
                if(constant == k) {
                    constant = 0;
                    sum++;
                }
            }
            if(sum >= m) {
                r = mid - 1;
            } else {
                l = mid + 1;
            }
        }
        return l;
    }
}

class Solution {
    public int splitArray(int[] nums, int m) {
        int n = nums.length;
        int l = 0, r = 0;
        for(int num : nums) {
            l = Math.max(l,num);
            r += num;
        }
        while(l <= r) {
            int mid = l + (r - l) / 2;
            int cnt = 1, sum = 0;
            for(int num : nums) {
                sum += num;
                if(sum <= mid) continue;
                cnt++; sum = num;
            }
            if(cnt > m) l = mid + 1;
            else r = mid - 1;
        }
        return l;
    }
}