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MaximumGap.java
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executable file
·71 lines (67 loc) · 2.45 KB
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/*
Author: King, wangjingui@outlook.com
Date: Dec 14, 2014
Problem: Maximum Gap
Difficulty: Hard
Source: https://oj.leetcode.com/problems/maximum-gap/
Notes:
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.
Solution: 1. Time : O(nlogn). Space : O(1);
Sort the unsorted array, and find the maximum difference.
2. Time : O(n). Space : O(n).
Drawer Theory. If we put n numbers into (n+1) drawers,
then there must be at least one empty drawer.
So we can find the maximum difference between two succesive non-empty drawers.
*/
public class Solution {
public int maximumGap_1(int[] num) {
Arrays.sort(num);
int res = 0;
for (int i = 1; i < num.length; ++i) {
res = Math.max(res, num[i] - num[i - 1]);
}
return res;
}
class node {
public int low;
public int high;
public node() {
low = -1;
high = -1;
}
}
public int maximumGap_2(int[] num) {
int n = num.length;
if (n < 2) return 0;
int minVal = num[0], maxVal = num[0];
for (int i = 1; i < n; ++i) {
minVal = Math.min(minVal, num[i]);
maxVal = Math.max(maxVal, num[i]);
}
//delta = (maxVal + 1 - minVal) / (n + 1)
//idx = (val - minVal) / delta = (val - minVal) * (n + 1) / (maxVal + 1 - minVal)
node[] pool = new node[n+2];
for (int i = 0; i < n+2; ++i) pool[i] = new node();
for (int i = 0; i < n; ++i) {
int idx =(int)(Long.valueOf(num[i] - minVal)* Long.valueOf(n + 1) / Long.valueOf(maxVal + 1 - minVal));
if (pool[idx].low == -1) {
pool[idx].low = pool[idx].high = num[i];
} else {
pool[idx].low = Math.min(pool[idx].low, num[i]);
pool[idx].high = Math.max(pool[idx].high, num[i]);
}
}
int pre = pool[0].high;
int res = 0;
for (int i = 1; i < n + 2; ++i) {
if (pool[i].low != -1) {
res = Math.max(res, pool[i].low - pre);
pre = pool[i].high;
}
}
return res;
}
}