leetcode 875. 爱吃香蕉的珂珂
珂珂喜欢吃香蕉。这里有 n 堆香蕉,第 i 堆中有 piles[i] 根香蕉。警卫已经离开了,将在 h 小时后回来。
珂珂可以决定她吃香蕉的速度 k (单位:根/小时)。每个小时,她将会选择一堆香蕉,从中吃掉 k 根。如果这堆香蕉少于 k 根,她将吃掉这堆的所有香蕉,然后这一小时内不会再吃更多的香蕉。
珂珂喜欢慢慢吃,但仍然想在警卫回来前吃掉所有的香蕉。
返回她可以在 h 小时内吃掉所有香蕉的最小速度 k(k 为整数)。
示例 1:
输入:piles = [3,6,7,11], h = 8
输出:4
示例 2:
输入:piles = [30,11,23,4,20], h = 5
输出:30
示例 3:
输入:piles = [30,11,23,4,20], h = 6
输出:23
提示:
1 <= piles.length <= 104
piles.length <= h <= 109
1 <= piles[i] <= 109
思路:使用二分法,取值从1-N,防止溢出,mid = (low+high)/2可以转换为int mid = (high - low) / 2 + low; 对pile/speed向上取整可以转换为(pile + speed - 1) / speed;注意total+的范围为long.
二分实现更新:
class Solution { public: long GetMax(vector<int> &piles, int k) { long time = 0; for (auto i : piles) { time += (i + k - 1) / k; } return time; } int GetMaxId(vector<int> &piles, int h) { int low = 1; int high = 0; for (auto i : piles) { high = max(i, high); } int res = high; while (low <= high) { int mid = (high - low) / 2 + low; long timeCnt = GetMax(piles, mid); if (timeCnt > h) { low = mid + 1; } else { high = mid - 1; res = mid; } } return res; } int minEatingSpeed(vector<int> &piles, int h) { return GetMaxId(piles, h); } };
代码:
#include <iostream> #include <map> #include <vector> #include <algorithm> #include <math.h> using namespace std; long GetMax(vector<int> &piles, int k) { long time = 0; for (auto i : piles) { time += ceil(i / (k * 1.0)); // (pile + speed - 1) / speed; } return time; } int GetMaxId(vector<int> &piles, int h) { int low = 1; int high = 0; for (auto i : piles) { high = max(i, high); } int res = high; while (low < high) { int mid = (high + low) / 2; //int speed = (high - low) / 2 + low; long timeCnt = GetMax(piles, mid); if (timeCnt > h) { low = mid + 1; } else { high = mid; res = mid; } } return res; } int minEatingSpeed(vector<int> &piles, int h) { return GetMaxId(piles, h); } int main() { vector<int> piles = { 312884470 }; int h = 968709470; int res = minEatingSpeed(piles, h); cout << "result: " << res << endl; return 0; }
以大多数人努力程度之低,根本轮不到去拼天赋~