实际二分搜索（写出函数，再用二分搜索法找左右边界

看到最大值的最小化，左边界，最小化的最大值，右边界

画图理解

爱吃香蕉的珂珂

class Solution {
    public int minEatingSpeed(int[] piles, int h) {
        int left=1,right=1000000000;
        while(left<=right){
            int mid=(left+right)/2;
            long hour=f(mid,piles);
            if(hour>h){
                left=mid+1;
            }else if(hour<=h){
                right=mid-1;
            }
        }
        return left;
    }
    long f(int x,int[] piles){
        long hour=0;
        for(int i=0;i<piles.length;i++){
            hour+=(piles[i]+x-1)/x;//向上取整方法
        }
        return hour;
    }
}

运货要通过数组的货物进行left和right的初始化

Left为数组最大值，right为总和，还要注意函数如何些，装货的逻辑

class Solution {
    public int shipWithinDays(int[] weights, int days) {
        int left=0,right=0;
        for(int x:weights){//初始化左右，left为数组最大值，right为数组总和
            if(x>left){
                left=x;
            }
            right+=x;
        }
        while(left<=right){
            int mid=(left+right)/2;
            long daysNeed=f(mid,weights);
                if(daysNeed>days){
                    left=mid+1;
                }else{
                    right=mid-1;
                }
            }
            return left;
    }
    long f(int x,int[] weights){
        long daysNeed=0;
        for(int i=0;i<weights.length;){
            int cap=x;//一次能装多少
            while(i<weights.length){//装货
                if(cap<weights[i]) break;
                else cap-=weights[i];
                i++;
            }//装满一次
            daysNeed++;
        }
        return daysNeed;
    }
}

class Solution {
    public int splitArray(int[] nums, int k) {
        int left=0,right=0;
        for(int x:nums){
            if(x>left) left=x;
            right+=x;
        }
        while(left<=right){
            int mid=(left+right)/2;
            int count=f(mid,nums);
            if(count<=k){
                right=mid-1;
            }else{
                left=mid+1;
            }
        }
        return left;
    }
    int f(int x,int[] nums){
        int count=0;
        for(int i=0;i<nums.length;){
            int cap=x;
            while(i<nums.length){
                if(cap<nums[i]) break;
                else{
                    cap-=nums[i];
                    i++;
                }
            }
            count++;
        }
        return count;
    }
}