LeetCode 2016. Maximum Difference Between Increasing Elements

原题链接在这里：https://leetcode.com/problems/maximum-difference-between-increasing-elements/

题目：

Given a 0-indexed integer array nums of size n, find the maximum difference between nums[i] and nums[j] (i.e., nums[j] - nums[i]), such that 0 <= i < j < n and nums[i] < nums[j].

Return the maximum difference. If no such i and j exists, return -1.

Example 1:

Input: nums = [7,1,5,4]
Output: 4
Explanation:
The maximum difference occurs with i = 1 and j = 2, nums[j] - nums[i] = 5 - 1 = 4.
Note that with i = 1 and j = 0, the difference nums[j] - nums[i] = 7 - 1 = 6, but i > j, so it is not valid.

Example 2:

Input: nums = [9,4,3,2]
Output: -1
Explanation:
There is no i and j such that i < j and nums[i] < nums[j].

Example 3:

Input: nums = [1,5,2,10]
Output: 9
Explanation:
The maximum difference occurs with i = 0 and j = 3, nums[j] - nums[i] = 10 - 1 = 9. 

Constraints:

• n == nums.length
• 2 <= n <= 1000
• 1 <= nums[i] <= 109

题解：

Update the min when encountering num in nums.

Update maxDiff with num - min.

Note: If nums keep decreasing or equal, then maxDiff should be -1. 

If there are equal in decreasing, then res = 0, thus return res == 0 ? -1 : res.

Time Complexity: O(n). n = nums.length.

Space: O(1).

AC Java:

 1 class Solution {
 2     public int maximumDifference(int[] nums) {
 3         if(nums == null || nums.length < 2){
 4             return -1;
 5         }
 6         
 7         int n = nums.length;
 8         int res = -1;
 9         int min = nums[0];
10         for(int i = 1; i < n; i++){
11             res = Math.max(res, nums[i] - min);
12             min = Math.min(min, nums[i]);
13         }
14         
15         return res == 0 ? -1 : res;
16     }
17 }

类似Best Time to Buy and Sell Stock.

标签：Difference,min,int,res,nums,Maximum,maximum,difference,2016
来源： https://www.cnblogs.com/Dylan-Java-NYC/p/16437202.html