复杂读分析

分析一下这段代码的时间复杂度：

int cal(int n) {
   int sum = 0;
   int i = 1;
   for (; i <= n; ++i) {
     sum = sum + i;
   }
   return sum;
}

每一行代码的执行时间是 1 ，for 循环执行了n次。所以是2+n次，去掉常数就是O(n)

int cal(int n) {
   int sum = 0;
   int i = 1;
   int j = 1;
   for (; i <= n; ++i) {
     j = 1;
     for (; j <= n; ++j) {
       sum = sum + i * j;
     }
   }
}

这段代码是俩个for循环。所以是O（n2）

int cal(int n) {
   int sum_1 = 0;
   int p = 1;
   for (; p < 100; ++p) {
     sum_1 = sum_1 + p;
   }

   int sum_2 = 0;
   int q = 1;
   for (; q < n; ++q) {
     sum_2 = sum_2 + q;
   }

   int sum_3 = 0;
   int i = 1;
   int j = 1;
   for (; i <= n; ++i) {
     j = 1;
     for (; j <= n; ++j) {
       sum_3 = sum_3 + i * j;
     }
   }

   return sum_1 + sum_2 + sum_3;
}

相加，复杂度是最大的那个

int cal(int n) {
   int ret = 0;
   int i = 1;
   for (; i < n; ++i) {
     ret = ret + f(i);
   }
}

int f(int n) {
int sum = 0;
int i = 1;
for (; i < n; ++i) {
    sum = sum + i;
}
return sum;
}

相乘的时间复杂度是O（n3）

i=1;
while (i <= n) {
i = i * 2;
}

i=1;
while (i <= n) {
i = i * 3;
}

时间复杂度是O(logn)

int cal(int m, int n) {
int sum_1 = 0;
int i = 1;
for (; i < m; ++i) {
sum_1 = sum_1 + i;
}

int sum_2 = 0;
int j = 1;
for (; j < n; ++j) {
sum_2 = sum_2 + j;
}

return sum_1 + sum_2;
}

时间复杂度是O(n)+O(n)

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