LintCode 183: Wood Cut (Binary Search 经典题!)

最新推荐文章于 2023-11-28 18:15:47 发布

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探讨了在给定数量的木材和所需小块数量下，如何通过算法找到能够切割出最多相同长度小块的最长长度。使用贪婪加二分查找策略，确保时间复杂度为O(nlogLen)。

Wood Cut
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.

思路：这题跟copy book那题很相似。最小值最大化，贪婪+二分。
注意最小值可以从1开始，最大值则可选木头的最大长度。

class Solution {
public:
    /**
     * @param L: Given n pieces of wood with length L[i]
     * @param k: An integer
     * @return: The maximum length of the small pieces
     */
    int woodCut(vector<int> &L, int k) {
        int len = L.size();
        if (len == 0) return 0;
        //int minLen = INT_MAX, maxLen = 0;
        int maxLen = 0;
        for (int i = 0; i < len; ++i) {
//            minLen = min(minLen, L[i]);
            maxLen = max(maxLen, L[i]);
        }
        
        int start = 1, end = maxLen;
        
        while (start + 1 < end) {
            int mid = start + (end - start) / 2;
            if (qualify(L, k, mid)) {
                start = mid;
            } else {
                end = mid;
            }
        }
        
        if (qualify(L, k, end)) return end;
        if (qualify(L, k, start)) return start;
        return 0;
    }
    
private:
    bool qualify(vector<int> &L, int k, int l) {
        int len = L.size();
        int count = 0;
        for (int i = 0; i < len; ++i) {
            count += L[i] / l;
        }
        return count >= k;
    } 
};