No.164 Maximum Gap

No.164 Maximum Gap

Given an unsorted array, find the maximum difference between the successive elements in its sorted form.

Try to solve it in linear time/space.

Return 0 if the array contains less than 2 elements.

You may assume all elements in the array are non-negative integers and fit in the 32-bit signed integer range.

题意理解：successive连续的
 　　　　输入：未排序数组，数组元素均为非负整数，在32位整数范围内
　　　　输出：输出排序后连续元素的最大差值
　　　　要求：线性时间/空间
法一：直观做法就是排序，之后一次比较，但时间空间效率不高!!!

class Solution
{
public:
    int    maximumGap(vector<int> &nums)
    {//题意理解：successive连续的
     //输入：未排序数组，数组元素均为非负整数，在32位整数范围内
     //输出：输出排序后连续元素的最大差值
     //要求：线性时间/空间
     //直观做法就是排序，之后一次比较，但时间空间效率不高!!!
        int count = nums.size();
        if(count <2)
            return 0;
        sort(nums.begin(),nums.end());
        int maxGap = nums[1] - nums[0];

        for(int i=2; i<count; i++)
        {
            if(maxGap < nums[i]-nums[i-1])
                maxGap = nums[i]-nums[i-1];
        }
        return maxGap;
    }
};

法二：

官方提示： 表示没有理解！！！

Suppose there are N elements and they range from A to B.

Then the maximum gap will be no smaller than ceiling[(B - A) / (N - 1)]

Let the length of a bucket to be len = ceiling[(B - A) / (N - 1)], then we will have at most num = (B - A) / len + 1 of bucket

for any number K in the array, we can easily find out which bucket it belongs by calculating loc = (K - A) / len and therefore maintain the maximum and minimum elements in each bucket.

Since the maximum difference between elements in the same buckets will be at most len - 1, so the final answer will not be taken from two elements in the same buckets.

For each non-empty buckets p, find the next non-empty buckets q, then q.min - p.max could be the potential answer to the question. Return the maximum of all those values.

 

 

参考：http://www.cnblogs.com/ganganloveu/p/4162290.html