[LeetCode] 1749. Maximum Absolute Sum of Any Subarray

 轉自LeetCode

You are given an integer array nums. The absolute sum of a subarray [numsl, numsl+1, ..., numsr-1, numsr] is abs(numsl + numsl+1 + ... + numsr-1 + numsr).

Return the maximum absolute sum of any (possibly empty) subarray of nums.

Note that abs(x) is defined as follows:

• If x is a negative integer, then abs(x) = -x.
• If x is a non-negative integer, then abs(x) = x.

 

Example 1:

Input: nums = [1,-3,2,3,-4]
Output: 5
Explanation: The subarray [2,3] has absolute sum = abs(2+3) = abs(5) = 5.

Example 2:

Input: nums = [2,-5,1,-4,3,-2]
Output: 8
Explanation: The subarray [-5,1,-4] has absolute sum = abs(-5+1-4) = abs(-8) = 8.

 

Constraints:

• 1 <= nums.length <= 105
• -104 <= nums[i] <= 104

Solution

想法滿簡單的

先求出最大和，和最小和，最後再都取絕對值來比大小

kotlin

	class Solution {
	fun maxAbsoluteSum(nums: IntArray): Int {
	var maxSum = nums[0]
	var minSum = nums[0]
	var tmpMaxSum = nums[0]
	var tmpMinSum = nums[0]
	for(i in 1..nums.lastIndex) {
	tmpMaxSum = Math.max(tmpMaxSum + nums[i], nums[i])
	tmpMinSum = Math.min(tmpMinSum + nums[i], nums[i])
	maxSum = Math.max(maxSum, tmpMaxSum)
	minSum = Math.min(minSum, tmpMinSum)
	}
	return Math.max(Math.abs(maxSum), Math.abs(minSum))
	}
	}

view raw maximum_absolute_sum_of_any_subarray.kt hosted with ❤ by GitHub