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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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。
时间O(n)
空间O(1)
Java实现
1 class Solution { 2 public int maximumDifference(int[] nums) { 3 int res = -1; 4 int min = Integer.MAX_VALUE; 5 for (int num : nums) { 6 res = Math.max(res, num - min); 7 min = Math.min(min, num); 8 } 9 return res > 0 ? res : -1; 10 } 11 }
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