面试题四十二：连续子数组的最大和，要求时间复杂度为 n

 1  int FindMaxArray(int [] A) {
 2               if(A==null||A.length<=0)
 3                   return 0;
 4               int sum=0;
 5               int max=0x80000000;//先定义为极小值-2147483648
 6               System.out.println(max);
 7               for(int i=0;i<A.length;++i) {
 8                   if(sum<=0) sum=A[i];
 9                   else sum+=A[i];
10                   if(sum>max) max=sum;            
11               }
12               return max;       
13          }