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;}
}
)
)
