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描述
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.
Note:
- n == nums.length
- 2 <= n <= 1000
- 1 <= nums[i] <= 10^9
解析
根据题意，给出了一个从 0 开始索引长度为 n 的列表 nums ，找出 nums[j] - nums[i] 的最大的差值，且 0 <= i < j < n ，且 nums[i] < nums[j] 。如果存在最大差值就返回，否则返回 -1 。
最简单的方法就是暴力双重循环：
- 初始化结果 result 为 -1
- 遍历 range(len(nums)-1) 中的每个元素 i ，再遍历 range(i+1, len(nums)) 中的每个元素 j ，如果 nums[j] <= nums[i] 则继续进行下一循环，否则用 result 记录 nums[j]-nums[i] 的最大值
- 遍历结束返回 result
解答
class Solution(object):
def maximumDifference(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
result = -1
for i in range(len(nums)-1):
for j in range(i+1, len(nums)):
if nums[j] <= nums[i]:
continue
result = max( result , nums[j]- nums[i])
return result
运行结果
Runtime: 316 ms, faster than 16.50% of Python online submissions for Maximum Difference Between Increasing Elements.
Memory Usage: 13.6 MB, less than 20.57% of Python online submissions for Maximum Difference Between Increasing Elements.
解析
也可以遍历一次就找到最大的差值，，思路：
- 初始化结果 result 为 -1 ，minNum 为 nums[0] 表示当前的最小值
- 从第二个元素开始，左到右每遍历到一个元素，如果 nums[i] 大于当前最小值 minNum ，就执行 max(result, nums[i]-minNum) 更新 result ，然后执行 min(minNum, nums[i]) 更新当前的最小值 minNum
- 遍历结束返回 result 即可
解答
class Solution(object):
def maximumDifference(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
result = -1
minNum = nums[0]
for i in range(1, len(nums)):
if nums[i]>minNum:
result = max(result, nums[i]-minNum)
minNum = min(minNum, nums[i])
return result
运行结果
Runtime: 36 ms, faster than 60.29% of Python online submissions for Maximum Difference Between Increasing Elements.
Memory Usage: 13.6 MB, less than 20.57% of Python online submissions for Maximum Difference Between Increasing Elements.
原题链接：leetcode.com/problems/ma…
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