Maximum Subarray - LeetCode

最新推荐文章于 2022-04-01 23:10:53 发布

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本文介绍了一种寻找含至少一个数的最大子数组和的算法。通过示例[-2,1,-3,4,-1,2,1,-5,4]说明了如何找到最大和为6的连续子数组[4,-1,2,1]。提供了两种解决方案：一种是O(n)的时间复杂度方法，另一种是使用分治策略的方法。

Find the contiguous subarray within an array (containing at least one number) which has the largest sum.

For example, given the array [−2,1,−3,4,−1,2,1,−5,4],
the contiguous subarray [4,−1,2,1] has the largest sum = 6.

click to show more practice.

More practice:

If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle.

要注意这里返回条件的界定和设定，十分容易出错！！！

这里的报错和我的Eclipse上显示的不一样，我认为我的方法应该是AC的，不知道是什么bug。

Input:	[-1,-2]
Output:	-2
Expected:	-1

public class Solution {
	 public int maxSubArray(int[] A) {
         int max = -9999;
     int maxindex = 0;
     int min = 9999;
     int minindex = 0;
     int sum = 0;
     int[] B = A;
     
     for(int i = 0; i < A.length; i++){
         sum = sum + A[i];
         B[i] = sum;
         //System.out.println(B[i]);
     }
     for(int i = 0; i < A.length; i++){
         
         maxindex = (B[i] > max)? i:maxindex;
         max = Math.max(B[i], max);
      //   System.out.println(maxindex);
     }
     for(int i = 0; i < A.length; i++){
         minindex = (B[i] < min)? i:minindex;
         min = Math.min(B[i], min);
     }
    // System.out.println(maxindex);
   //  System.out.println(minindex);
     if(maxindex == minindex)
     return A[maxindex];
     else{
         if(maxindex > minindex){
             return B[maxindex] - B[minindex];
         }
         else
             return - B[maxindex] + B[minindex];
     }
     
     }
}

给一个我喜欢的AC方法：

public class Solution {
	public int maxSubArray(int[] A) {
		int max = A[0];
		int[] sum = new int[A.length];
		sum[0] = A[0];
 
		for (int i = 1; i < A.length; i++) {
			sum[i] = Math.max(A[i], sum[i - 1] + A[i]);
			max = Math.max(max, sum[i]);
		}
 
		return max;
	}
}