[LeetCode] 2016. Maximum Difference Between Increasing Elements

轉自LeetCode

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

Solution

這題其實和 121. Best Time to Buy and Sell Stock 是同個題目

題目規定 i < j，也是一個要先買後賣的概念

那要注意的是，題目還有規定 nums[i] < nums[j]

所以在找最小值的時候，檢查條件要變成 nums[i] <= min

kotlin

	class Solution {
	fun maximumDifference(nums: IntArray): Int {
	var min = Int.MAX_VALUE
	var ans = -1
	for(n in nums) {
	if(n <= min) {
	min = n
	} else {
	ans = Math.max(ans, n - min)
	}
	}
	return ans
	}
	}

view raw maximum_difference_between_increasing_elements.kt hosted with ❤ by GitHub