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LintCode-Copy Books
Given an array A of integer with size of n( means n books and number of pages of each book) and k people to copy the 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.) Each person have can copy one page per minute. Return the number of smallest minutes need to copy all the books.
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例子
Given array A = [3,2,4], k = 2.
Return 5( First person spends 5 minutes to copy book 1 and book 2 and second person spends 4 minutes to copy book 3. )
挑战
Could you do this in O(n*k) time ?
分析：仅仅会O(n*lgM),M为全部数的总和，能够看到，求最小要多少分钟。也就是大于这个数，都是能够的。而小于这个数则都是不能够的。于是能够利用二分来找到这个点
代码：
class Solution {
public:
/**
* @param pages: a vector of integers
* @param k: an integer
* @return: an integer
*/
int copyBooks(vector<int> &pages, int k) {
// write your code here
int l = 0;
int r = 99999999;
while(l<r)
{
int mid = (l+r)/2;
if(check(mid,pages,k))
{
r = mid;
}
else
l = mid+1;
}
return l;
}
bool check(int index,vector<int>& pages,int k)
{
int cnt = 0;
int sum = 0;
int i = 0;
while(i<pages.size())
{
if(sum+pages[i]<=index)
{
sum+=pages[i++];
}
else if(pages[i]<=index)
{
cnt++;
sum = 0;
}
else
return false;
}
if(sum!=0)
cnt++;
return cnt<=k;
}
};
LintCode-Copy Books
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