45题 Maximum Subarray Difference最大子数组差_maximum subarray difference 分析-优快云博客

本文链接：https://blog.youkuaiyun.com/caomengying/article/details/117172460

本文介绍了一种解决最大子数组差问题的算法，通过Java实现，针对给定的整数数组，寻找两个非重叠子数组，使得它们和的绝对差值最大。算法利用了四个数组来分别记录每个位置的最大和最小前缀和，最后遍历计算可能的子数组差值并返回最大值。

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Maximum Subarray Difference 最大子数组差

Description
Given an array with integers.

Find two non-overlapping subarrays A and B, which |SUM(A) - SUM(B)|∣SUM(A)−SUM(B)∣ is the largest.

Return the largest difference.

public class Solution {
    /**
     * @param nums: A list of integers
     * @return: An integer indicate the value of maximum difference between two substrings
     */
    public int maxDiffSubArrays(int[] nums) {
        // write your code here
        if(nums == null || nums.length == 0){
          return 0 ;
        }
        int size = nums.length ;
        int[] left_max = new int[size] ;
        int[] left_min = new int[size] ;
        int[] right_max = new int[size] ;
        int[] right_min = new int[size] ;
        int[] copy = new int[size] ;
        for(int i = 0 ; i < size ; i++){
          copy[i] = -1 * nums[i] ;
        }
        int max = Integer.MIN_VALUE, sum = 0 , minSum = 0 ;
        for(int i = 0 ; i < nums.length ; i++){
               sum = sum + nums[i] ;
               max = Math.max(max , sum - minSum) ;
               minSum = Math.min(sum , minSum) ;
               left_max[i] = max ;
        }
         max = Integer.MIN_VALUE ; 
         sum = 0 ; 
         minSum = 0 ;
        for(int i = size-1 ; i >= 0 ; i--){
               sum = sum + nums[i] ;
               max = Math.max(max , sum - minSum) ;
               minSum = Math.min(sum , minSum) ;
               right_max[i] = max ;
        }
        max = Integer.MIN_VALUE ; 
         sum = 0 ; 
         minSum = 0 ;
        for(int i = 0 ; i < size ; i++){
           sum += copy[i] ;
           max = Math.max(max , sum-minSum ) ;
           minSum = Math.min(minSum , sum) ;
           left_min[i] = -1 * max ;
        }
        max = Integer.MIN_VALUE ; 
         sum = 0 ; 
         minSum = 0 ;
        for(int i = size-1 ; i >= 0 ; i--){
           sum += copy[i] ;
           max = Math.max(max , sum-minSum ) ;
           minSum = Math.min(minSum , sum) ;
           right_min[i] = -1 * max ;
        }
        int diff =0 ;
        for(int i= 0 ; i< size -1 ; i++){
          diff = Math.max(diff, Math.abs(left_max[i] - right_min[i+1]));
          diff = Math.max(diff, Math.abs(left_min[i] - right_max[i+1]));
        }
        return diff ;
    }
}