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183. Wood Cut

• 本题难度: Hard

• Topic: Binary Search

  Description

  Given n pieces of wood with length L[i] (integer array). Cut them into small pieces to guarantee you could have equal or more than k pieces with the same length. What is the longest length you can get from the n pieces of wood? Given L & k, return the maximum length of the small pieces.

Example
For L=[232, 124, 456], k=7, return 114.

Challenge
O(n log Len), where Len is the longest length of the wood.

Notice
You couldn‘t cut wood into float length.

If you couldn‘t get >= k pieces, return 0.

我的代码

class Solution:
    """
    @param L: Given n pieces of wood with length L[i]
    @param k: An integer
    @return: The maximum length of the small pieces
    """
    def woodCut(self, L, k):
        # write your code here
     # write your code here
        if L == []:
            return 0
        L.sort()
        max = L[-1]
        l = 1
        r = L[-1]
        while(l<=r):
            count = 0
            m = (l + r) // 2
            print(l,r,m)
            for i in L:
                count += i//m
            if count>=k:
                l = m+1
            else:
                r = m-1
        #1.边界问题，最后结果可能是从右向左逼近，此时m为第一个不满足条件的情况，反之，从左往右逼近，则m为满足条件的情况
        print(m)
        count = 0
        for i in L:
            count += i//m
        print(count)
        if count >= k:
            return m
        else:
            return m-1

思路

1. 切割的大小肯定小于最长木料长度，而从最长木料选择截取特定长的的时间复杂度是log(L)
2. 共有n块木料，每个特定的长度要试n次

• 时间复杂度: nlog(L) L为最长木料长度
• 出错
  想得挺顺利的，就是一个边界问题没考虑好，最后结果可能是从右向左逼近，此时m为第一个不满足条件的情况，反之，从左往右逼近，则m为满足条件的情况
  也有低级错误

[Lintcode]183. Wood Cut

标签：@param   while   des   amp   14.   pie   nlog   for   oat   

原文地址：https://www.cnblogs.com/siriusli/p/10358600.html