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【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 };
【leetcode】Maximum Subarray
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