分割数组的子数组和的最大值最小问题（二分法+贪心算法）

以上两个题考察内容一模一样，均是将数组拆分成一定数量的子数组，求子数组和的最大值，并求解这个最大值的底线为多少？
实现思路：二分法查找

C++版本

class Solution {
     
public:
    int shipWithinDays(vector<int>& weights, int D) {
     
        // 确定二分查找左右边界
        int left = *max_element(weights.begin(), weights.end()), right = accumulate(weights.begin(), weights.end(), 0);
        while (left < right) {
     
            int mid = (left + right) / 2;
            // need 为需要运送的天数
            // cur 为当前这一天已经运送的包裹重量之和
            int need = 1, cur = 0;
            for (int weight: weights) {
     
                if (cur + weight > mid) {
     
                    ++need;
                    cur = 0;
                }
                cur += weight;
            }
            if (need <= D) {
     
                right = mid;
            }
            else {
     
                left = mid + 1;
            }
        }
        return left;
    }
};

java版本

class Solution {
     
    public int splitArray(int[] nums, int m) {
     
     int left=Arrays.stream(nums).max().getAsInt(),right=Arrays.stream(nums).sum();
     while(left<right){
     
         int mid=(right+left)/2;
         int need=1,cur=0;
         for(int weight:nums){
     
             if(cur+weight>mid){
     
                 need++;
                 cur=0;
             }
             cur+=weight;
         }
         if(need<=m){
     
           right=mid;
         }
         else{
     
             left=mid+1;
         }
     }
     return left;
    }
}

再比如这个吃香蕉问题：

class Solution {
     
    public int minEatingSpeed(int[] piles, int h) {
     
     int left=1,right=Arrays.stream(piles).max().getAsInt();
     while(left<right){
     
         int mid=(right+left)/2;//mid=K
         int need=0;
         for(int weight:piles){
         
             if(weight<=mid){
     
                 need++;         
             }
             else{
     
               need+=(weight+mid-1)/mid;  
             }      
         }
         if(need<=h){
     
           right=mid;
         }
         else{
     
             left=mid+1;
         }
     }
     return left;
    }
}