LeetCode 1749. Maximum Absolute Sum of Any Subarray

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

解析：子数组和的绝对值最大。

采用前缀和的思路，分别计算前缀和最小的数据，前缀和最大的数值。

class Solution {
public:
    int maxAbsoluteSum(vector<int>& nums) {
        int len = nums.size();
        vector<int> pre_sum(len + 1, 0);
        pre_sum[0] = 0;
        for (int i = 1; i <= len; i++) {
            pre_sum[i] = pre_sum[i - 1] + nums[i - 1];
        }
        int max_pre = 0, min_pre = 0;
        for (int i = 1; i < pre_sum.size(); i++) {
            max_pre = max(max_pre, pre_sum[i]);
            min_pre = min(min_pre, pre_sum[i]);
        }
        return abs(max_pre - min_pre);
    }
};