C++ 最大子序列和的时间复杂度计算
大家好,我正在计算最大子序列和的时间复杂度。 实际上我知道答案是O(n^3),它是从函数(n^3+3n^2+2n)/6得出的C++ 最大子序列和的时间复杂度计算,c++,big-o,time-complexity,C++,Big O,Time Complexity,大家好,我正在计算最大子序列和的时间复杂度。 实际上我知道答案是O(n^3),它是从函数(n^3+3n^2+2n)/6得出的 我的问题是该函数是如何获得的。实际上非常简单:只需查看代码中的循环即可 for (int i=0; i<n; i++) for(j = i; j<n; j++) { ... for (int k=i; k<=j; k++) XXX; for(inti=0;i实际上非常简单:只需查看代码中的
我的问题是该函数是如何获得的。实际上非常简单:只需查看代码中的循环即可
for (int i=0; i<n; i++)
for(j = i; j<n; j++) {
...
for (int k=i; k<=j; k++)
XXX;
for(inti=0;i实际上非常简单:只需查看代码中的循环即可
for (int i=0; i<n; i++)
for(j = i; j<n; j++) {
...
for (int k=i; k<=j; k++)
XXX;
for(inti=0;i实际上非常简单:只需查看代码中的循环即可
for (int i=0; i<n; i++)
for(j = i; j<n; j++) {
...
for (int k=i; k<=j; k++)
XXX;
for(inti=0;i实际上非常简单:只需查看代码中的循环即可
for (int i=0; i<n; i++)
for(j = i; j<n; j++) {
...
for (int k=i; k<=j; k++)
XXX;
for(inti=0;i以下是如何
i=0
j=0 k=0 (count=1 )
j=1 k=0,1 (count =2)
j=2 k=0,1,2 (count = 3)
...
j=n-1 k=0,1,2,...n-1 (count = n)
Total number of times code executed = 1+2+3+...+n = n(n+1)/2
i=1
j=1 k=1 (count=1 )
j=2 k=1,2 (count =2)
j=3 k=1,2, 3 (count = 3)
...
j=n-1 k=1,2,...n-1 (count = n-2)
Total number of times code executed = 1+2+3+...+n-1 = (n-1)n/2
...
i=n-1
j=n-1 k=n-1 ( count = 1)
Total number of of times code executed = 1 = 1(1+1)/2
Now if we sum for all the values of i
n(n+1)/2 + ((n-1)((n-1)+1)/2+.....+1(1+1)/2
=∑ N(N+1)/2 =1/2∑(N^2 +N) =1/2(∑N^2+∑N)=1/2{ 1/6 N(N+1)(2N+1) + 1/2 N(N+1) } =1/2{ (2N^3 + 3N^2+N )/6 +(N^2+N)/2} =(N^3 + 3N^2 + 2N)/6
下面是如何
i=0
j=0 k=0 (count=1 )
j=1 k=0,1 (count =2)
j=2 k=0,1,2 (count = 3)
...
j=n-1 k=0,1,2,...n-1 (count = n)
Total number of times code executed = 1+2+3+...+n = n(n+1)/2
i=1
j=1 k=1 (count=1 )
j=2 k=1,2 (count =2)
j=3 k=1,2, 3 (count = 3)
...
j=n-1 k=1,2,...n-1 (count = n-2)
Total number of times code executed = 1+2+3+...+n-1 = (n-1)n/2
...
i=n-1
j=n-1 k=n-1 ( count = 1)
Total number of of times code executed = 1 = 1(1+1)/2
Now if we sum for all the values of i
n(n+1)/2 + ((n-1)((n-1)+1)/2+.....+1(1+1)/2
=∑ N(N+1)/2 =1/2∑(N^2 +N) =1/2(∑N^2+∑N)=1/2{ 1/6 N(N+1)(2N+1) + 1/2 N(N+1) } =1/2{ (2N^3 + 3N^2+N )/6 +(N^2+N)/2} =(N^3 + 3N^2 + 2N)/6
下面是如何
i=0
j=0 k=0 (count=1 )
j=1 k=0,1 (count =2)
j=2 k=0,1,2 (count = 3)
...
j=n-1 k=0,1,2,...n-1 (count = n)
Total number of times code executed = 1+2+3+...+n = n(n+1)/2
i=1
j=1 k=1 (count=1 )
j=2 k=1,2 (count =2)
j=3 k=1,2, 3 (count = 3)
...
j=n-1 k=1,2,...n-1 (count = n-2)
Total number of times code executed = 1+2+3+...+n-1 = (n-1)n/2
...
i=n-1
j=n-1 k=n-1 ( count = 1)
Total number of of times code executed = 1 = 1(1+1)/2
Now if we sum for all the values of i
n(n+1)/2 + ((n-1)((n-1)+1)/2+.....+1(1+1)/2
=∑ N(N+1)/2 =1/2∑(N^2 +N) =1/2(∑N^2+∑N)=1/2{ 1/6 N(N+1)(2N+1) + 1/2 N(N+1) } =1/2{ (2N^3 + 3N^2+N )/6 +(N^2+N)/2} =(N^3 + 3N^2 + 2N)/6
下面是如何
i=0
j=0 k=0 (count=1 )
j=1 k=0,1 (count =2)
j=2 k=0,1,2 (count = 3)
...
j=n-1 k=0,1,2,...n-1 (count = n)
Total number of times code executed = 1+2+3+...+n = n(n+1)/2
i=1
j=1 k=1 (count=1 )
j=2 k=1,2 (count =2)
j=3 k=1,2, 3 (count = 3)
...
j=n-1 k=1,2,...n-1 (count = n-2)
Total number of times code executed = 1+2+3+...+n-1 = (n-1)n/2
...
i=n-1
j=n-1 k=n-1 ( count = 1)
Total number of of times code executed = 1 = 1(1+1)/2
Now if we sum for all the values of i
n(n+1)/2 + ((n-1)((n-1)+1)/2+.....+1(1+1)/2
=∑ N(N+1)/2 =1/2∑(N^2 +N) =1/2(∑N^2+∑N)=1/2{ 1/6 N(N+1)(2N+1) + 1/2 N(N+1) } =1/2{ (2N^3 + 3N^2+N )/6 +(N^2+N)/2} =(N^3 + 3N^2 + 2N)/6
马克·艾伦·韦斯(在他的书中)建议的检查。马克·艾伦·韦斯(在他的书中)建议的检查。马克·艾伦·韦斯(在他的书中)建议的检查。马克·艾伦·韦斯(在他的书中)建议的检查