如何比较两个字符串如果它们被重新排列

Question

我想知道如何比较两个重新排列的字符串例如，如果字符串a =“字符串”，字符串b =“tsrngi”...如果我比较a.equals（b），它将返回false，因为字符的顺序。我希望它返回true，因为字符相同但只是顺序不同..谢谢

Answer 1

Sort them, then compare. To sort, use something like:

char[] content = unsorted.toCharArray();
java.util.Arrays.sort(content);
String sorted = new String(content);

Answer 2

I really like JRL's solution, since it's quite elegant. At the same time, I feel that because there is a solution that is an order of complexity better that I should share it. It's less elegant, but it's O(n) instead of O(n lg n) .

if(str1.length() != str2.length()) return false; 
Map<Character, Integer> counts = new HashMap<Character, Integer>();

for(int i = 0; i < str1.length(); i++) {
    // add 1 for count for str1
    if(counts.contains(str1.charAt(i)) {
        counts.put(str1.charAt(i),counts.get(star1.charAt(i)) + 1);
    } else {
        counts.put(str1.charAt(i),1);
    }
    // sub 1 for count for str2
    if(counts.contains(str1.charAt(i)) {
        counts.put(str1.charAt(i),counts.get(star1.charAt(i)) - 1);
    } else {
        counts.put(str1.charAt(i),-1);
    }
}
// when you're done, all values in the map should be 0. If they
// aren't all 0, you don't have equal-arranged strings.
for(Integer i : counts.values()) {
    if(i.intValue() != 0) return false;
}
// we made it this far, we know it's true
return true;