Java的String类中length()函数的复杂度是多少? [英] What is complexity of length() function in String class of Java?

问题描述

是 O(n) 还是 O(1)(通过在向对象分配字符串期间将长度保存在私有变量中).

Is it O(n) or O(1) (By saving length in a private variable during string allocation to the object).

如果是O(n)，是不是意味着下面代码的复杂度是O(n^2)?

if it is O(n), does it mean that the complexity of following code is O(n^2)?

for(int i=0; i<s.length()-1;i++){
    //some code here!
}

推荐答案

O(1) 因为String 实例的长度是已知的.

It is O(1) as the length is already known to String instance.

从 JDK 1.6 开始可见.

From JDK 1.6 it is visible.

public int length() {
    return count;
}

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更新

了解为什么他们可以缓存 count 的值并继续为 count 使用相同的值很重要.原因在于他们在设计 String 时做出了一个伟大的决定，即它的不可变性.

It is important to understand why they can cache the value of count and keep using same value for count. The reason lies in a great decision they took when designing String, its Immutability.

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