time-complexity

来源： https://stackoverflow.com/questions/59797066/why-the-time-complexity-of-the-following-c-code-is-onn

来源： https://stackoverflow.com/questions/59797066/why-the-time-complexity-of-the-following-c-code-is-onn

问题 This question already has answers here : Why is time complexity for BFS/DFS not simply O(E) instead of O(E+V)? (2 answers) Breadth First Search time complexity analysis (6 answers) Closed 9 days ago . This is a generic BFS implementation: For a connected graph with V nodes and E total number of edges, we know that every edge will be considered twice in the inner loop. So if the total number of iterations in the inner loop of BFS is going to be 2 * number of edges E , isn't the runtime going

问题 This question already has answers here : Why is time complexity for BFS/DFS not simply O(E) instead of O(E+V)? (2 answers) Breadth First Search time complexity analysis (6 answers) Closed 9 days ago . This is a generic BFS implementation: For a connected graph with V nodes and E total number of edges, we know that every edge will be considered twice in the inner loop. So if the total number of iterations in the inner loop of BFS is going to be 2 * number of edges E , isn't the runtime going

问题 I know there's a similar question in stack overflow, where one person has asked, why time complexity of BFS/DFS is not simply O(V). The appropriate answer given was that E can be as large as V^2 in case of complete graph, and hence it is valid to include E in time complexity. But, if V cannot be greater than E+1. So, in that case not having V in the time complexity, should work? 回答1: If it is given that E = kV + c , for some real constants k and c then, O(E + V) = O(kV + c + V) = O(V) = O(E)

问题 I know there's a similar question in stack overflow, where one person has asked, why time complexity of BFS/DFS is not simply O(V). The appropriate answer given was that E can be as large as V^2 in case of complete graph, and hence it is valid to include E in time complexity. But, if V cannot be greater than E+1. So, in that case not having V in the time complexity, should work? 回答1: If it is given that E = kV + c , for some real constants k and c then, O(E + V) = O(kV + c + V) = O(V) = O(E)

问题 I have 2 lists of tuples list1 = [(1.332, 3.23344, 3.22), (2.122, 2.11, 2.33), ... (1, 2, 3)] and list2 = [(4.23, 12.2, 3.333), (1.234, 3.21, 4.342), ... (1.1, 2.2, 3.3)] . These lists are both very long, somewhere in the millions for both lists. For context, each of these data points is some measure of position in two different datasets. Now I want to correspond each entry in list1 to an entry in list2 if it is "close enough". By close enough I mean the distance between the positions is less

问题 I just wonder the complexity of this regex matching problem: given a string of small letters and a matching rule, determine whether the rule may match the WHOLE string. The rule is a simplified regex which only contains smaller letters and/or '.' (period) and/or '*' (asterisk). A period may match any small letter where an asterisk may match zero or more of the preceding element. Here are some examples: isMatch("aa","a") is false isMatch("aa","aa") is true isMatch("aaa","aa") is false isMatch(

问题 I just wonder the complexity of this regex matching problem: given a string of small letters and a matching rule, determine whether the rule may match the WHOLE string. The rule is a simplified regex which only contains smaller letters and/or '.' (period) and/or '*' (asterisk). A period may match any small letter where an asterisk may match zero or more of the preceding element. Here are some examples: isMatch("aa","a") is false isMatch("aa","aa") is true isMatch("aaa","aa") is false isMatch(

问题 How may I find the time complexity of the following code: (Sorry for adding image, I will re-edit my question once I have access to the laptop) What I have done so far: The first loop iterates n times, the second i times and the third log(i*j) times, So after simplifying I got: Sigma from i=0 to n for i*log i + n * (Sigma from j=0 to i for log i) But why this is equal to O(n^2 log(n))? 回答1: If we look at the outermost 2 loops we observe that there are 1 + 2 + 3 ... + n - 1 iterations. Using