Yyds dry inventory solution sword finger offer: maximum sum of continuous subarrays (II)

1. sketch ：

describe

Enter a length of n Integer array array, One or more consecutive integers in an array form a subarray , Find a continuous subarray with the largest sum .1. Subarray is continuous , such as [1,3,5,7,9] The subarrays of are [1,3],[3,5,7] wait , however [1,3,7] It's not a subarray 2. If there are multiple consecutive subarrays with the largest sum , Then return the longest , The data of this question ensure that there is only one 3. The minimum length of the subarray defined by this question is 1, There is no empty subarray , That there is no [] Is a sub array of an array 4. The returned array is not included in the space complexity calculation

Data range :

requirement : Time complexity , Spatial complexity Advanced : Time complexity , Spatial complexity

Example 1

Input ：

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       [1,-2,3,10,-4,7,2,-5]
      
      
      
      

       
       
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Return value ：

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       [3,10,-4,7,2]
      
      
      
      

       
       
1.
      
      

explain ：

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        It can be seen from the analysis , Subarray of input array [3,10,-4,7,2] The maximum sum can be obtained as 18, So go back to [3,10,-4,7,2]
      
      
      
      

       
       
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Example 2

Input ：

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       [1]
      
      
      
      

       
       
1.
      
      

Return value ：

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       [1]
      
      
      
      

       
       
1.
      
      

Example 3

Input ：

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       [1,2,-3,4,-1,1,-3,2]
      
      
      
      

       
       
1.
      
      

Return value ：

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       [1,2,-3,4,-1,1]
      
      
      
      

       
       
1.
      
      

explain ：

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        It can be seen from the analysis , The sum of the largest subarray is 4, Yes [4],[4,-1,1],[1,2,-3,4],[1,2,-3,4,-1,1], Therefore, the longest one is returned [1,2,-3,4,-1,1]
      
      
      
      

       
       
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Example 4

Input ：

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       [-2,-1]
      
      
      
      

       
       
1.
      
      

Return value ：

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       [-1]
      
      
      
      

       
       
1.
      
      

explain ：

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        The minimum length of the subarray is 1, So go back to [-1]
      
      
      
      

       
       
1.
      
      

2. Code implementation ：

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       import java.util.*;
       
       
public class Solution {
       
       
    public int[] FindGreatestSumOfSubArray (int[] array) {
       
       
        int x = array[0];
       
       
        int y = 0;
       
       
        int maxsum = x;
       
       
        // Sliding interval 
       
       
        int left = 0, right = 0; 
       
       
        // Record the longest interval 
       
       
        int resl = 0, resr = 0; 
       
       
        for(int i = 1; i < array.length; i++){
       
       
            right++;
       
       
            // State shift ： Continuous subarray and maximum 
       
       
            y = Math.max(x + array[i], array[i]); 
       
       
            // New starting point of the interval 
       
       
            if(x + array[i] < array[i]) 
       
       
                left = right;
       
       
            // Update Max 
       
       
            if(y > maxsum || y == maxsum && (right - left + 1) > (resr - resl + 1)){ 
       
       
                maxsum = y;
       
       
                resl = left;
       
       
                resr = right;
       
       
            }
       
       
            // to update x The state of 
       
       
            x = y; 
       
       
        }
       
       
        // Take an array 
       
       
        int[] res = new int[resr - resl + 1];
       
       
        for(int i = resl; i <= resr; i++) 
       
       
            res[i - resl] = array[i];
       
       
        return res;
       
       
    }
       
       
}
       
       

      
      
      
      

       
       
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