【LeetCode】164. Maximum Gap (2 solutions)

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.

Credits:
Special thanks to @porker2008 for adding this problem and creating all test cases.

解法一：

先排序O(nlogn)，再一次遍历，得到maxGap

虽然不满足O(n)的时间要求，但是最直观的想法。

class Solution {
public:
    int maximumGap(vector<int>& nums) {
        if(nums.empty() || nums.size() == )
            return ;
        sort(nums.begin(), nums.end());
        int ret = ;
        for(int i = ; i < nums.size(); i ++)
            ret = max(ret, nums[i]-nums[i-]);
        return ret;
    }
};

解法二：为了满足O(n)复杂度，我尝试了计数排序，但是会TLE。因此使用桶排序来做。

(计数排序可以看做是桶大小为1的桶排序，但由于桶数目太多导致遍历时间过长。)

最大gap肯定是出现在后一个有效桶的min与前一个有效桶的max之间。

“有效”指的是忽略空桶。

class Solution {
public:
    int maximumGap(vector<int>& nums) {
        if(nums.empty() || nums.size() == )
            return ;
        int n = nums.size();
        int minAll = *min_element(nums.begin(), nums.end());
        int maxAll = *max_element(nums.begin(), nums.end());
        // type conversion!!!
        double gap = ((double)(maxAll - minAll)) / (n - );
        // compute min and max element for each bucket
        vector<int> minV(n-, INT_MAX);
        vector<int> maxV(n-, INT_MIN);
        for(int i = ; i < n; i ++)
        {
            if(nums[i] != maxAll)
            {// the bktId of maxAll will fall out of bucket range
                int bktId = (int)((nums[i]-minAll)/gap);
                minV[bktId] = min(minV[bktId], nums[i]);
                maxV[bktId] = max(maxV[bktId], nums[i]);
            }
        }
        int ret = ;
        int curMax = maxV[];
        for(int i = ; i < n-; i ++)
        {
            if(minV[i] != INT_MAX)
            {
                ret = max(ret, minV[i]-curMax);
                curMax = maxV[i];
            }
        }
        ret = max(ret, maxAll-curMax);
        return ret;
    }
};