[LintCode] 183. 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.

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The unit of length is centimeter.The length of the woods are all positive integers,you couldn't cut wood into float length.If you couldn't get >= k pieces, return 0.

Example 1

Input:
L = [232, 124, 456]
k = 7
Output: 114
Explanation: We can cut it into 7 pieces if any piece is 114 long, however we can't cut it into 7 pieces if any piece is 115 long.
And for the 124 logs, the excess can be discarded and not used in its entirety.

Example 2

Input:
L = [1, 2, 3]
k = 7
Output: 0
Explanation: It is obvious we can't make it.

Challenge

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

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。给定L和k，你需要计算能够得到的小段木头的最大长度。

这道题的思路是二分，而且是在答案上二分。对于所有的木头而言，切割的长度的下界是1，上界是最长的那个木材的长度。在这个范围内进行二分，用每次二分得到的潜在的mid长度再去计算到底能切割成几个小段。因为想使得每段长度尽可能更长，所以 33 行我们先 return right。

时间O(nlogn)

空间O(1)

Java实现

public 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
     */
    public int woodCut(int[] logs, int k) {
        // write your code here
        int len = logs.length;
        long sum = 0;
        int max = 0;
        for (int log : logs) {
            max = Math.max(max, log);
            sum += log;
        }
        // corner case
        if (sum < k) {
            return 0;
        }

        // normal case
        int left = 1;
        int right = max;
        while (left + 1 < right) {
            int mid = left + (right - left) / 2;
            if (isValid(logs, k, mid)) {
                left = mid;
            } else {
                right = mid;
            }
        }
        if (isValid(logs, k, right)) {
            return right;
        }
        return left;
    }

    private boolean isValid(int[] logs, int k, int mid) {
        int count = 0;
        for (int log : logs) {
            count += log / mid;
        }
        if (count >= k) {
            return true;
        }
        return false;
    }
}

 

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