Question

我试图分析这段代码的时间复杂程度，但是我被卡住了。我是否认为它是O（n ^ 2）时间复杂度，因为有两个for循环，或者它只是O（n），因为第二个for循环并不总是运行？使用此代码，我必须基于图形

扫描二维数组

//adj is edgeMatrix
public int[] getClosenessCentrality(int[][] adj){
    int size = adj.length * adj.length;
    int[] closeness = new int[size];
    for (int vertex = 0; vertex < adj.length; vertex++) {
        boolean[] visited = new boolean[size];
        PriorityQueue<Integer> pq = new PriorityQueue<Integer>(size);


        pq.add(vertex);


        while (!pq.isEmpty()) {
            int u = pq.remove();
            if(!visited[u]) {
                visited[u] = true;
                for (int i = 0; i < size; i++) {
                    //Priority Queue speeds up extract-min
                    if (!visited[i]) {
                        if (adj[u][i] > 0) {
                            pq.add(adj[u][i]+1);
                        } 

                    }
                }
                closeness[vertex] += 1;
            }

        }
    }

    return closeness;
}

Answer 1

如果n = adj.length的复杂性为O(n^3 log(n))。

这就是原因。

n vertex n^2个O(log(n^2)) = O(2 log(n)) = O(log(n))的不同值：
对于所有O(n^3)可能的边缘，我们：
添加到优先级队列，并从优先级队列中删除。（这些都是O(log(n))次操作。）

所以把它们放在一起，我们O(n^3 log(n))总共有let textToType = "This is what I want you to type."; const typedLetters = []; document.getElementById('user-input').addEventListener("keypress", function(event) { const key = event.which || event.keyCode; const nextLetter = textToType[0].charCodeAt(); const outputTarget = document.getElementById("display-text"); const greenWrapper = document.createElement("span"); greenWrapper.classList.add("typed-cursor"); if (key === nextLetter) { typedLetters.push(String.fromCharCode(nextLetter)); textToType = textToType.substr(1); greenWrapper.textContent = typedLetters.join(""); outputTarget.textContent = textToType; outputTarget.prepend(greenWrapper); }; })个body { font-family: monospace; } .title-of-page>h1 { text-align: center; font-family: monospace; } .title-of-page { background-color: #414a4c; color: #ced3db; } .jumbotron { margin: 0; } .navigation-bar { background-color: #46494f; } a { color: green; } .nav>li>a:hover { background-color: #878f9b; } .navbar-nav>li { text-align: center; float: none; display: table-cell; } .navbar-nav { display: table; width: 100%; margin: 0; } .navbar { margin: 0; padding: 0; border-radius: 0; } .typing-field { width: 60em; height: 8em; background-color: #7e7e7f; opacity: 0.4; margin-left: 15em; margin-top: 5em; border: 3px solid black; padding: 0.8em; } #display-text { color: white; font-size: 2em; } .user-input { font-size: 1em; padding-left: 35em; padding-top: 2em; } #timer { padding-top: 4em; padding-left: 10em; font-size: 1.5em; color: red; } .typed-cursor { opacity: 1; -webkit-animation: blink 0.7s infinite; -moz-animation: blink 0.7s infinite; animation: blink 0.7s infinite; color: black; } @keyframes blink { 0% { opacity: 1; } 50% { opacity: 0; } 100% { opacity: 1; } } @-webkit-keyframes blink { 0% { opacity: 1; } 50% { opacity: 0; } 100% { opacity: 1; } } @-moz-keyframes blink { 0% { opacity: 1; } 50% { opacity: 0; } 100% { opacity: 1; } }可能的复杂<div class="jumbotron title-of-page container-fluid"> <h1>Typing Counter</h1> </div> <nav class="navbar navigation-bar container-fluid"> <div class=""> <ul class="nav navbar-nav"> <li><a href="#">Home</a></li> <li><a href="#">Contest</a></li> <li><a href="#">About</a></li> <li><a href="#">Leaderboard</a></li> </ul> </div> </nav> <div> <div id="timer"> <button type="button" class="btn" onclick="timer();">Start</button> </div> <div class="typing-field"> <p id="display-text">This is what I want you to type. </div> <div class="user-input"> <input type="text" name="user-input-text-box" id="user-input" /> </div> </div>操作。

如何详细计算此代码的复杂度大O？

1 个答案: