[LeetCode] 2016. Maximum Difference Between Increasing Elements

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

增量元素之间的最大差值。

给你一个下标从 0 开始的整数数组 nums ，该数组的大小为 n ，请你计算 nums[j] - nums[i] 能求得的 最大差值 ，其中 0 <= i < j < n 且 nums[i] < nums[j] 。

返回 最大差值 。如果不存在满足要求的 i 和 j ，返回 -1 。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/maximum-difference-between-increasing-elements
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思路

这道题就是股票题版本一换了个说法。唯一需要注意的是如果不存在满足题意的差值，返回的是 -1。

复杂度

时间O(n)
空间O(1)

代码

Java实现

class Solution {
    public int maximumDifference(int[] nums) {
        int res = -1;
        int min = Integer.MAX_VALUE;
        for (int num : nums) {
            res = Math.max(res, num - min);
            min = Math.min(min, num);
        }
        return res > 0 ? res : -1;
    }
}

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