在 JavaScript 中比较字符串的最佳方法？
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Optimum way to compare strings in JavaScript?
提问by HRJ
I am trying to optimize a function which does binary search of strings in JavaScript.
我正在尝试优化一个在 JavaScript 中对字符串进行二进制搜索的函数。
Binary search requires you to know whether the key is ==the pivot or <the pivot.
二分查找要求您知道键是==枢轴还是<枢轴。
But this requires two string comparisons in JavaScript, unlike in Clike languages which have the strcmp()function that returns three values (-1, 0, +1)for (less than, equal, greater than).
但这需要 JavaScript 中的两个字符串比较，这与C具有strcmp()返回三个值(-1, 0, +1)（小于、等于、大于）的函数的语言不同。
Is there such a native function in JavaScript, that can return a ternary value so that just one comparison is required in each iteration of the binary search?
JavaScript 中是否有这样一个本机函数，它可以返回一个三元值，以便在二进制搜索的每次迭代中只需要进行一次比较？
回答by Daniel Vassallo
You can use the localeCompare()method.
您可以使用该localeCompare()方法。
string_a.localeCompare(string_b);
/* Expected Returns:
0: exact match
-1: string_a < string_b
1: string_a > string_b
*/
Further Reading:
进一步阅读：
回答by Cipi
Well in JavaScript you can check two strings for values same as integers so yo can do this:
好吧，在 JavaScript 中，您可以检查两个字符串中是否有与整数相同的值，因此您可以这样做：
"A" < "B""A" == "B""A" > "B"
"A" < "B""A" == "B""A" > "B"
And therefore you can make your own function that checks strings the same way as the strcmp().
因此，您可以创建自己的函数，以与strcmp().
So this would be the function that does the same:
所以这将是执行相同操作的函数：
function strcmp(a, b)
{
return (a<b?-1:(a>b?1:0));
}
回答by Gumbo
You can use the comparison operators to compare strings. A strcmpfunction could be defined like this:
您可以使用比较运算符来比较字符串。一个strcmp函数可以这样定义：
function strcmp(a, b) {
if (a.toString() < b.toString()) return -1;
if (a.toString() > b.toString()) return 1;
return 0;
}
Edit????Here's a string comparison function that takes at most min { length(a), length(b) } comparisons to tell how two strings relate to each other:
编辑????这里的一个字符串比较函数，它至多分钟{长度（一），长度（b）}的比较告诉两个字符串如何彼此相关：
function strcmp(a, b) {
a = a.toString(), b = b.toString();
for (var i=0,n=Math.max(a.length, b.length); i<n && a.charAt(i) === b.charAt(i); ++i);
if (i === n) return 0;
return a.charAt(i) > b.charAt(i) ? -1 : 1;
}
