力扣labuladong——一刷day26

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二分搜索的泛化问题
首先从问题中抽象出自变量X，就是题目要求取的最值
其次，抽象出映射函数F(X)，捋清对应关系，使其单调
最后，计算F(X) == target，返回X

前言

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一、力扣875. 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-left)/2;
            if(fun(mid,piles) == h){
                right = mid - 1;
            }else if(fun(mid,piles) > h){
                left = mid + 1;
            }else{
                right = mid - 1;
            }
        }
        return left;
    }
    public long fun(int k, int[] piles){
        long hour = 0;
        for(int i = 0; i < piles.length; i ++){
            hour += piles[i]/k;
            if(piles[i]%k > 0){
                hour ++;
            }
        }
        return hour;
    }
}

二、力扣1011. 在 D 天内送达包裹的能力

class Solution {
    public int shipWithinDays(int[] weights, int days) {
        int left = 0, right = 0;
        for(int i = 0; i < weights.length; i ++){
            left = Math.max(left,weights[i]);
            right += weights[i];
        }
        while(left <= right){
            int mid = left + (right - left)/2;
            if(fun(mid, weights) == days){
                right = mid - 1;
            }else if(fun(mid, weights) > days){
                left = mid + 1;
            }else{
                right = mid - 1;
            }
        }
        return left;
    }
    public int fun(int k, int[] weights){
        int day = 0;
        for(int i = 0; i < weights.length;){
            int cap = k;
            while(i < weights.length){
                if(cap < weights[i])break;
                else cap -= weights[i++];
            }
            day ++;
        }
        return day;
    }
}

三、力扣410. 分割数组的最大值

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