如何将嵌套的for循环复杂性降低到python中的单个循环?

本文介绍了如何将嵌套的for循环复杂性降低到python中的单个循环?的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述 for i in range(0,x): for j in range(0,y): if (i+j)%2 == 0:

想想像同时掷两个骰子，然后发现骰子上的总和是否为偶数，但这里有一个问题，一个骰子有 6 个面，但这里两个面可以有任意数量的大小，相等和不相等甚至相等！任何人都可以建议如何在一个循环下合并它，因为我想不出任何一个?

Think of something like tossing two dices at the same time and finding if the sum on the dices is an even number but here's the catch, a dice has 6 sides but here the two can have any number of sizes, equal and not equal even! Can anyone suggest how to merge it under one loop because I can't think of any?

推荐答案

你无法摆脱嵌套循环(你可以隐藏它，比如使用 itertool.product，但它会仍然在某处执行，复杂度仍然是 O(x * y)) 但是你可以摆脱条件，如果你只需要生成满足它的 j 的值，通过调整 j 的范围.

You can't get rid of the nested loop (you could hide it, like by using itertool.product, but it would still be executed somewhere, and the complexity would still be O(x * y)) but you can get rid of the condition, if you only need to generate the values of j that satisfy it, by adapting the range for j.

这样，通过避免无用的循环，您将减少大约两倍的循环.

This way, you'll have about twice as less loops by avoiding the useless ones.

for i in range(0,x): for j in range(i%2,y, 2): print(i, j, i+j)

输出:

0 0 0 0 2 2 1 1 2 1 3 4 2 0 2 2 2 4