使用hashcode（）和length（）进行字符串比较(String comparison using hashcode() and length())

 在Java（或任何其他PL使用相同的哈希码功能）是否可以安全地说： 
 
 
 
 两个相等长度的字符串具有相同的散列码当且仅当它们相等时 
 
 
 让我们假设hashcode函数将会是 
 
 
 
 s [0] * 31 ^（n-1）+ s [1] * 31 ^（n-2）+ ... + s [n-1] 
 
 
 用例： 
 
 为了比较两个大规模的字符串集合，使用option 1而不是option 2在理论上会更快，因为哈希码计算将会执行一次，因为字符串将缓存它的值： 
 
 选项1： 
 
 for(String s1 : collection1){
   for(String s2 : collection2){
     if((s1.hashCode() == s2.hashCode()) && (s1.length()==s2.length()){
        System.out.println("matched");
     }
   }
 }
 
 方案2 
 
 for(String s1 : collection1){
   for(String s2 : collection2){
     if(s1.equals(s2)){
        System.out.println("matched");
     }
   }
 }
 
 更新： 
 
 在@tobias_k发表评论后，我意识到这个说法错了，所以我改变了这个问题。 
 
 
 
 是否有一个字符串的最大长度为M，对于任何两个长度相等的字符串，它们的哈希码将相同，当且仅当它们相等时 
 

In java (or any other PL that uses same hashcode function) Is it safe to say that:
 
 
 
Two strings of equal length have same hashcode if and only if they are equal 
 
 
let's just assume the hashcode function will be 
 
 
 
s[0]*31^(n-1) + s[1]*31^(n-2) + ... + s[n-1]
 
 
use case:
 
for comparing two massive collections of strings, it'll be theoretically faster to use option 1 rather than option 2 because hashcode computation will be done once as String will cache the its value:
 
OPTION 1:
 
 for(String s1 : collection1){
   for(String s2 : collection2){
     if((s1.hashCode() == s2.hashCode()) && (s1.length()==s2.length()){
        System.out.println("matched");
     }
   }
 }
 
OPTION 2
 
 for(String s1 : collection1){
   for(String s2 : collection2){
     if(s1.equals(s2)){
        System.out.println("matched");
     }
   }
 }
 
UPDATE:
 
after @tobias_k comment I realized that statment wrong, So I change the question.
 
 
 
is there a max length M for string that for any two strings of equal length their hashcode will be same if and only if they are equal
 

原文：https://stackoverflow.com/questions/31027909