0001 int sumI ( int A[], int n ) { //数组求和算法(迭代版) 0002 int sum = 0; //初始化累计器,O(1) 0003 for ( int i = 0; i < n; i++ ) //对全部共O(n)个元素,逐一 0004 sum += A[i]; //累计,O(1) 0005 return sum; //返回累计值,O(1) 0006 } //O(1) + O(n)*O(1) + O(1) = O(n+2) = O(n)