int maxSubArray(int* nums, int numsSize){ if(numsSize==0) return 0; if(numsSize==1) return nums[0]; return helper(nums,0,numsSize-1); } int helper(int *nums,int l,int r) { int mid=(l+r)/2; if(l>=r) return nums[l]; int left=helper(nums,l,mid-1); int right=helper(nums,mid+1,r); int lsum=0,rsum=0; int max1=0; int max2=0; int max; for(int i=mid-1;i>=l;i--) { lsum+=nums[i]; max1=max1>lsum?max1:lsum; } for(int j=mid+1;j<=r;j++) { rsum+=nums[j]; max2=max2>rsum?max2:rsum; } max=left>right?left:right; return max>(max1+max2+nums[mid])?max:(max1+max2+nums[mid]); }
①考虑中间元素时,左右两边序列的最大序列和最小应为0
原文:https://www.cnblogs.com/KIROsola/p/13595522.html