leetcode164.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.

解法：

桶排序，首先找到数组的最大值和最小值，然后设置gap为最大值和最小值的差距除以数组长度，那么我们所需要找到的max值则可以通过遍历每个桶，求当前桶的最大值和下一个桶的最小值的差距，这个差距的最大值即为所求。

代码：

class Solution {
public:
    int maximumGap(vector<int>& nums) {
        if(nums.size() < 2)
            return 0;
        if(nums.size() == 2)
            return abs(nums[0]-nums[1]);
        int maxnum = nums[0];
        int minnum = nums[0];
        for(int i = 0; i < nums.size(); i ++)
        {
            maxnum = max(maxnum, nums[i]);
            minnum = min(minnum, nums[i]);
        }

        int gap = ceil((double)(maxnum-minnum)/nums.size());
        if(gap == 0)
            return 0;
        int len = (maxnum-minnum)/gap+1;
        int maxi[len];
        int mini[len];

        for(int i = 0; i < len; i ++)
        {
            maxi[i] = minnum;
            mini[i] = maxnum;
        }

        int rank = 0;
        for(int i = 0; i < nums.size(); i ++)
        {
            rank = (nums[i] - minnum) / gap;
            maxi[rank] = max(maxi[rank],nums[i]);
            mini[rank] = min(mini[rank],nums[i]);
        }

        int ans = 0;
        int tempmax = maxi[0];
        for(int i = 1; i < len; i ++)
        {
            if(maxi[i] == minnum && mini[i] == maxnum)
                continue;
            else
            {
                ans = max(ans,(mini[i]-tempmax));
                tempmax = maxi[i];
            }
            // ans = max(ans, (mini[i]-maxi[i-1]));
        }
        return ans;
    }
};

注意：

1 gap可能为0
2 有的桶可能为空，所以不能简单的下一个桶的最小值减去当前桶的最大值。