下列程序中sumFun()的算法时间复杂度为 ( ) 。int sumFun(int n) { int count = 0; for (int i = 1; i < n; i = i * 2) for (int j = i; j > 0; j = j - 1) count = count + i + j; return count; } A: O(n) B: O(nlogn) C: O(n^2) D: O(logn logn)
下列程序中sumFun()的算法时间复杂度为 ( ) 。int sumFun(int n) { int count = 0; for (int i = 1; i < n; i = i * 2) for (int j = i; j > 0; j = j - 1) count = count + i + j; return count; } A: O(n) B: O(nlogn) C: O(n^2) D: O(logn logn)
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