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- Copy Books II
Given n books( the page number of each book is the same) and an array of integer with size k means k people to copy the book and the i th integer is the time i th person to copy one book). You must distribute the continuous id books to one people to copy. (You can give book A[1],A[2] to one people, but you cannot give book A[1], A[3] to one people, because book A[1] and A[3] is not continuous.) Return the number of smallest minutes need to copy all the books.
Example
Given n = 4, array A = [3,2,4], .
Return 4( First person spends 3 minutes to copy book 1, Second person spends 4 minutes to copy book 2 and 3, Third person spends 4 minutes to copy book 4. )
Solution:
- Greedy algorithm + Binary Search
这题跟437. Copy Books很象,思路也是贪婪+二分。不同之处在于437里面,每本书的厚度不一样,而抄写员速度一样,而这题书的厚度都一样,但抄写员抄写速度不一样。
这题也是问最大值的最小化。因为也是问的时间,所以二分的对象还是时间。但是贪婪算法里面checkValid(times, t, n) 变成了判断给定数组times,给定时间t,问能不能抄完n本书。注意437里面的checkValid(pages, t, k)是判断给定数组pages,给定时间t,问k个人能不能搞定。
另外,这题的二分的left和right的开始值可以根据抄写员的最慢和最快速度而定。
代码如下:
class Solution {
public:
/**
* @param n: An integer
* @param times: an array of integers
* @return: an integer
*/
int copyBooksII(int n, vector<int> ×) {
int k = times.size();
int totalTime = 0;
int minTime = INT_MAX, maxTime = 0;
for (int i = 0; i < k; ++i) {
minTime = min(minTime, times[i]);
maxTime = max(maxTime, times[i]);
}
int left = n * minTime / k;
int right = n * maxTime / k;
while (left + 1 < right) {
int mid = left + (right - left) / 2;
if (checkValid(times, mid, n)) {
right = mid;
} else {
left = mid;
}
}
if (checkValid(times, left, n)) return left;
else return right;
}
private:
//given array times, time t,
is it ok to finish n books in time?
bool checkValid(vector<int> ×, int t, int n) {
int copyTime = times[0];
int count = 0; //count of books
int k = times.size();
for (int i = 0; i < k; ++i) {
count += t / times[i]; // the ith guy can copy t/times[i] books
}
return count >= n;
}
};
//input cases:
//1) 100
//[1,2]
//2) 4
//[3,2,4]