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.

class Solution {    public int maximumGap(int[] nums) {        if (nums.length<2) return 0;        int minNum = -1, maxNum = -1, n = nums.length;        for (int i=0; i<n; ++i) {            minNum = min(nums[i], minNum);            maxNum = max(nums[i], maxNum);        }        if (maxNum==minNum) return 0;        double average = (maxNum-minNum) * 1.0 / (n - 1);        if (average==0) ++average;        int[] localMin = new int[n];        int[] localMax = new int[n];        for (int i=0; i<n; ++i) {            localMin[i] = -1;            localMax[i] = -1;        }        for (int i=0; i<n; ++i) {            int t = (int)((nums[i]-minNum) / average);            localMin[t] = min(localMin[t], nums[i]);            localMax[t] = max(localMax[t], nums[i]);        }        int ans = (int)average, left = 0, right = 1;        while (left<n-1) {            while (right<n && localMin[right]==-1) ++right;            if (right>=n) break;            ans = max(ans, localMin[right]-localMax[left]);            left = right;            ++right;        }        return ans;    }    private int min(int a, int b) {        if (a==-1) return b;        else             if (b==-1) return a;            else                if (a<b) return a;                else return b;    }    private int max(int a, int b) {        if (a==-1) return b;        else            if (b==-1) return a;            else                if (a>b) return a;                else return b;        }}