lintcode_183.木材加工

有一些原木，现在想把这些木头切割成一些长度相同的小段木头，需要得到的小段的数目至少为 k。当然，我们希望得到的小段越长越好，你需要计算能够得到的小段木头的最大长度。

 注意事项

木头长度的单位是厘米。原木的长度都是正整数，我们要求切割得到的小段木头的长度也要求是整数。无法切出要求至少 k 段的,则返回 0 即可。

样例

有3根木头[232, 124, 456], k=7, 最大长度为114.

 

问题分析：

要求满足段数条件下的最大长度，最大长度上界应该等于 sum(L) / k，当切的长度大于上界无论如何无法得到。

下界无要求，正整数即可，先用顺序搜索

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
        limit = sum(L) / k
        tmp = limit
        while True:
            num = 0

            if tmp < 1:
                return 0     
            for i in L:
                num += i / tmp
            if num >= k:
                return tmp
            tmp -= 1

也可以用二分查找，问题实质找到满足条件情况下的最大值

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
        limit = sum(L) / k

        start, end = 1, limit
        while start + 1 < end:
            middle = (start + end) / 2
            
            num = 0
            for i in L:
                num += i / middle
            if num >= k:
                start = middle
            else:
                end = middle
        
        num = 0
        for i in L:
            num += i / start
        if num >= k:
            return start
        else:
            return 0

九章算法:

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):
        if sum(L) < k:
            return 0
            
        maxLen = max(L)
        start, end = 1, maxLen
        while start + 1 < end:
            mid = (start + end) / 2
            pieces = sum([l / mid for l in L])
            if pieces >= k:
                start = mid
            else:
                end = mid
                
        if sum([l / end for l in L]) >= k:
            return end
        return start

写法更简洁，界限考虑不太一样