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Polish the chapter
introduction, computational complexity.
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@ -132,37 +132,37 @@ int main() {
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cout << "输入数据大小 n = " << n << endl;
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int count = constant(n);
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cout << "常数阶的计算操作数量 = " << count << endl;
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cout << "常数阶的操作数量 = " << count << endl;
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count = linear(n);
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cout << "线性阶的计算操作数量 = " << count << endl;
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cout << "线性阶的操作数量 = " << count << endl;
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vector<int> arr(n);
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count = arrayTraversal(arr);
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cout << "线性阶(遍历数组)的计算操作数量 = " << count << endl;
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cout << "线性阶(遍历数组)的操作数量 = " << count << endl;
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count = quadratic(n);
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cout << "平方阶的计算操作数量 = " << count << endl;
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cout << "平方阶的操作数量 = " << count << endl;
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vector<int> nums(n);
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for (int i = 0; i < n; i++)
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nums[i] = n - i; // [n,n-1,...,2,1]
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count = bubbleSort(nums);
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cout << "平方阶(冒泡排序)的计算操作数量 = " << count << endl;
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cout << "平方阶(冒泡排序)的操作数量 = " << count << endl;
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count = exponential(n);
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cout << "指数阶(循环实现)的计算操作数量 = " << count << endl;
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cout << "指数阶(循环实现)的操作数量 = " << count << endl;
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count = expRecur(n);
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cout << "指数阶(递归实现)的计算操作数量 = " << count << endl;
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cout << "指数阶(递归实现)的操作数量 = " << count << endl;
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count = logarithmic((float)n);
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cout << "对数阶(循环实现)的计算操作数量 = " << count << endl;
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cout << "对数阶(循环实现)的操作数量 = " << count << endl;
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count = logRecur((float)n);
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cout << "对数阶(递归实现)的计算操作数量 = " << count << endl;
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cout << "对数阶(递归实现)的操作数量 = " << count << endl;
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count = linearLogRecur((float)n);
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cout << "线性对数阶(递归实现)的计算操作数量 = " << count << endl;
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cout << "线性对数阶(递归实现)的操作数量 = " << count << endl;
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count = factorialRecur(n);
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cout << "阶乘阶(递归实现)的计算操作数量 = " << count << endl;
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cout << "阶乘阶(递归实现)的操作数量 = " << count << endl;
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return 0;
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}
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