找到字符串中的第一个非重复字符的最佳方式 [英] Best way to find first non repeating character in a string

问题描述

什么是最好的空间和时间，有效的解决方案，以找到像 aabccbdcbe ？

字符串的第一个非重复字符

这里的答案为d。所以这令我的一点是，它可以通过两种方式来完成：

1. 对于每一个指标i循环时，i-1次，检查是否曾经再次出现该字符。但是，这是没有效率：此方法的生长为O（N ^ 2）其中N是串的长度。
2. 另一种可能的好办法可能是，如果我能形成一个树或使得我可以命令字符基于所述权重（所述出现计数）任何其他的ds。这可能需要我长N只是一个遍历字符串组成的结构。这仅仅是O（N）+ O（时间来建立树或任何DS）。

解决方案

下面是一个非常简单的 O（N）解决方法：

  DEF FN（S）：
  为了= []
  计数= {}
  对于x在S：
    如果x的罪状：
      计数[X] + = 1
    其他：
      计数[X] = 1
      order.append（X）
  对于x依次是：
    如果计数[X] == 1：
      返回X
  返回None
 

通过串一旦我们循环。当我们遇到一个新的角色，我们将其存储在计数与 1 的值，并追加到序。当我们遇到我们以前见过一个角色，我们增加了在值计数。最后，我们通过序，直到我们找到 1 的值的字符计数循环返回。

What would be the best space and time efficient solution to find the first non repeating character for a string like aabccbdcbe?

The answer here is d. So the point that strikes me is that it can be done in two ways:

1. For every index i loop i-1 times and check if that character occurs ever again. But this is not efficient: growth of this method is O(N^2) where N is the length of the string.
2. Another possible good way could be if I could form a tree or any other ds such that I could order the character based on the weights (the occurrence count). This could take me just one loop of length N through the string to form the structure. That is just O(N) + O(time to build tree or any ds).

解决方案

Here's a very straightforward O(n) solution:

def fn(s):
  order = []
  counts = {}
  for x in s:
    if x in counts:
      counts[x] += 1
    else:
      counts[x] = 1 
      order.append(x)
  for x in order:
    if counts[x] == 1:
      return x
  return None

We loop through the string once. When we come across a new character, we store it in counts with a value of 1, and append it to order. When we come across a character we've seen before, we increment its value in counts. Finally, we loop through order until we find a character with a value of 1 in counts and return it.

这篇关于找到字符串中的第一个非重复字符的最佳方式的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！