【leetcode】Maximum Subarray

Maximum Subarray

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.

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.

 

在累加的过程中，如果发现sum<0则说明前面的序列对后面的序列没有贡献，故此时设置sum=0

 

 1 class Solution {

 2 public:

 3     int maxSubArray(int A[], int n) {

 4        

 5         int sum,maxSum;

 6         sum=maxSum=A[0];

 7         for(int i=1;i<n;i++)

 8         {

 9             if(sum<0) sum=0;

10             sum+=A[i];

11            

12             if(maxSum<sum) maxSum=sum;

13         }

14         return maxSum;       

15     }

16 };

 

采用分治法求解：

找到左半边最大的序列值，找到右半边最大的序列值，找到中间序列的值

 

 1 class Solution {

 2 public:

 3     int maxSubArray(int A[], int n) {

 4        

 5         divideAndConquer(A,0,n-1);

 6     }

 7    

 8     int divideAndConquer(int A[],int left,int right)

 9     {

10        

11         if(left==right) return A[left];

12  

13         int mid=(left+right)/2;

14        

15        

16         int leftMax=divideAndConquer(A,left,mid);

17         int rightMax=divideAndConquer(A,mid+1,right);

18        

19         int midSum1=0;

20         int midMax1=A[mid];

21        

22         for(int i=mid;i>=left;i--)

23         {

24             midSum1+=A[i];

25             if(midMax1<midSum1) midMax1=midSum1;

26         }

27        

28         int midSum2=0;

29         int midMax2=A[mid+1];

30        

31         for(int i=mid+1;i<=right;i++)

32         {

33             midSum2+=A[i];

34             if(midMax2<midSum2) midMax2=midSum2;

35         }

36        

37         int midMax=midMax1+midMax2;

38        

39         return max(max(leftMax,rightMax),midMax);

40        

41        

42     }

43 };