查找给定字符串的排名与重复所有可能的排列名单

问题描述

我试图找到排列的列表中指定字符串的排名，并希望有人能找到的bug。

I was trying to find the Rank of the given string in the list of permutations and was hoping someone can find the bug.

function permute() {
    var W = $('input').val(),
        C = [];
    for (var i = 0; i < 26; i++) C[i] = 0;
    var rank = 1;
    for (var i = 0; i < W.length; i++) {
        C[W.charCodeAt(i) - 'a'.charCodeAt(0)]++;
    }
    var repeated= 1;
    for (var i = 0; i < C.length; i++) {
        if(C[i] > 0) {
            repeated *=  fact(C[i]);
        }
    }    

    if (W !== '') {
        for (var i = 0; i < W.length; i++) {
            //How many characters which are not used, that come before current character
            var count = 0;
            for (var j = 0; j < 26; j++) {
                if (j == (W.charCodeAt(i) - 'a'.charCodeAt(0))) break;
                if (C[j] > 0) count++;
            }
            C[W.charCodeAt(i) - 'a'.charCodeAt(0)] = 0;
            rank += ( count * fact(W.length - i - 1) );
        }
        rank = rank/ repeated;
    }
    var pp = 'Rank of  :: ' + W + ' -- ' + rank;
    $('div').append('<p>' + pp + '</p>');
}

function fact(n) {
    if (n == 0 || n == 1) return 1;
    else return fact(n - 1) * n;
}

$('button').click(permute);

检查小提琴

Check Fiddle

一个用例，这可能是

会计是应该给的等级 10743

推荐答案

这里是演示：

对于每个位置上检查有多少留字有重复的，使用的逻辑，如果你需要重排ñ东西，如果'一'的东西是相似的排列数为 N！/年！

For each position check how many characters left have duplicates, and use the logic that if you need to permute n things and if 'a' things are similar the number of permutations is n!/a!

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