[LeetCode] Maximum Gap 求最大间距

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.

 

遇到这类问题肯定先想到的是要给数组排序，但是题目要求是要线性的时间和空间，那么只能用桶排序或者基排序。这里我用桶排序来做，首先找出数组的最大值和最小值，然后要确定每个桶的容量，即为(最大值 - 最小值) / 个数 + 1，在确定桶的个数，即为(最大值 - 最小值) / 桶的容量 + 1，然后需要在每个桶中找出局部最大值和最小值，而最大间距的两个数不会在同一个桶中，而是一个桶的最小值和另一个桶的最大值之间的间距。代码如下：

 

class Solution {
public:
    int maximumGap(vector<int> &num) {
        if (num.size() < 2) return 0;
        int n = num.size();
        int minN = num[0];
        int maxN = num[0];
        for (int i = 1; i < n; ++i) {
            maxN = max(num[i], maxN);
            minN = min(num[i], minN);
        }
        int len = (maxN - minN) / n + 1;
        int bucketNum = (maxN - minN) / len + 1;
        vector<int> bucketMin(bucketNum, INT_MAX);
        vector<int> bucketMax(bucketNum, INT_MIN);
        for (int i = 0; i < n; ++i) {
            int Idx = (num[i] - minN) / len;
            bucketMin[Idx] = min(bucketMin[Idx], num[i]);
            bucketMax[Idx] = max(bucketMax[Idx], num[i]);
        }
        int res = 0;
        int pre = 0;
        for (int i = 1; i < bucketNum; ++i) {
            if (bucketMin[i] == INT_MAX || bucketMax[i] == INT_MIN) continue;
            res = max(res, bucketMin[i] - bucketMax[pre]);
            pre = i;
        }
        return res;
    }
};

 

参考：

http://blog.csdn.net/u011345136/article/details/41963051

http://blog.csdn.net/u012162613/article/details/41936569

 

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