nested-loops | 易学教程

for loop vs while loop vs foreach loop PHP

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问题 1st off I\'m new to PHP. I have been using for loop,while loop,foreach loop in scripts. I wonder which one is better for performance? what\'s the criteria to select a loop? which should be used when we loop inside another loop? the code which I\'m stuck with wondering which loop to be used. for($i=0;$i<count($all);$i++) { //do some tasks here for($j=0;$j<count($rows);$j++) { //do some other tasks here } } It\'s pretty obvious that I can write the above code using while. Hope someone will help

How to break out of nested loops?

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问题 If I use a break statement, it will only break inner loop and I need to use some flag to break the outer loop. But if there are many nested loops, the code will not look good. Is there any other way to break all of the loops? (Please don\'t use goto stmt .) for(int i = 0; i < 1000; i++) { for(int j = 0; j < 1000; j++) { if(condition) { // both of the loops need to break and control will go to stmt2 } } } stmt2 回答1: Use: if (condition) { i = j = 1000; break; } 回答2: No, don't spoil the fun with

Iterate Multi-Dimensional Array with Nested Foreach Statement

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问题 I think this might be a pretty simple question, but I haven\'t been able to figure it out yet. If I\'ve got a 2-dimensional array like so: int[,] array = new int[2,3] { {1, 2, 3}, {4, 5, 6} }; What\'s the best way to iterate through each dimension of the array with a nested foreach statement? 回答1: If you want to iterate over every item in the array as if it were a flattened array, you can just do: foreach (int i in array) { Console.Write(i); } which would print 123456 If you want to be able

What is the Big-O of a nested loop, where number of iterations in the inner loop is determined by the current iteration of the outer loop?

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What is the Big-O time complexity of the following nested loops: for(int i = 0; i < N; i++) { for(int j = i + 1; j < N; j++) { System.out.println("i = " + i + " j = " + j); } } Would it be O(N^2) still? Alex Gaynor Yep, it's still O(n^2), it has a smaller constant factor, but that doesn't affect O notation. Yes. Recall the definition of Big-O: O(f(n)) by definition says that the run time T(n) ≤ kf(n) for some constant k . In this case, the number of steps will be (n-1)+(n-2)+...+0 , which rearranges to the sum of 0 to n-1; this is T(n)=(n-1)((n-1)+1)/2 . Rearrange that and you can see that T(n

What is the Big-O of a nested loop, where number of iterations in the inner loop is determined by the current iteration of the outer loop?

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问题 What is the Big-O time complexity of the following nested loops: for(int i = 0; i < N; i++) { for(int j = i + 1; j < N; j++) { System.out.println(\"i = \" + i + \" j = \" + j); } } Would it be O(N^2) still? 回答1: Yep, it's still O(n^2), it has a smaller constant factor, but that doesn't affect O notation. 回答2: Yes. Recall the definition of Big-O: O(f(n)) by definition says that the run time T(n) ≤ kf(n) for some constant k . In this case, the number of steps will be (n-1)+(n-2)+...+0 , which

Breaking out of a nested loop

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问题 If I have a for loop which is nested within another, how can I efficiently come out of both loops (inner and outer) in the quickest possible way? I don\'t want to have to use a boolean and then have to say go to another method, but rather just to execute the first line of code after the outer loop. What is a quick and nice way of going about this? Thanks I was thinking that exceptions aren\'t cheap/should only be thrown in a truly exceptional condition etc. Hence I don\'t think this solution

How to break nested loops in JavaScript? [duplicate]

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问题 This question already has an answer here: What's the best way to break from nested loops in JavaScript? 15 answers I tried this: for(i = 0; i < 5; i++){ for(j = i + 1; j < 5; j++){ break(2); } alert(1); } only to get: SyntaxError : missing ; before statement So, how would I break a nested loop in JavaScript? 回答1: You should be able to break to a label, like so: function foo () { dance: for(var k = 0; k < 4; k++){ for(var m = 0; m < 4; m++){ if(m == 2){ break dance; } } } } 回答2: You need to

What's the best way to break from nested loops in JavaScript?

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问题 What\'s the best way to break from nested loops in Javascript? //Write the links to the page. for (var x = 0; x < Args.length; x++) { for (var Heading in Navigation.Headings) { for (var Item in Navigation.Headings[Heading]) { if (Args[x] == Navigation.Headings[Heading][Item].Name) { document.write(\"<a href=\\\"\" + Navigation.Headings[Heading][Item].URL + \"\\\">\" + Navigation.Headings[Heading][Item].Name + \"</a> : \"); break; // <---HERE, I need to break out of two loops. } } } } 回答1:

Can I use break to exit multiple nested for loops?

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问题 Is it possible to use the break function to exit several nested for loops? If so, how would you go about doing this? Can you also control how many loops the break exits? 回答1: AFAIK, C++ doesn't support naming loops, like Java and other languages do. You can use a goto, or create a flag value that you use. At the end of each loop check the flag value. If it is set to true, then you can break out of that iteration. 回答2: No, don't spoil it with a break . This is the last remaining stronghold for

What's the best way to break from nested loops in JavaScript?

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What's the best way to break from nested loops in Javascript? //Write the links to the page. for (var x = 0; x < Args.length; x++) { for (var Heading in Navigation.Headings) { for (var Item in Navigation.Headings[Heading]) { if (Args[x] == Navigation.Headings[Heading][Item].Name) { document.write("<a href=\"" + Navigation.Headings[Heading][Item].URL + "\">" + Navigation.Headings[Heading][Item].Name + "</a> : "); break; // <---HERE, I need to break out of two loops. } } } } ephemient Just like Perl, loop1: for (var i in set1) { loop2: for (var j in set2) { loop3: for (var k in set3) { break