levenshtein.html

<pre>
<script>
function levenshtein(a,b) {
    var table = [[[]]];
    

    
    for(var j = 1; j <= b.length; j++) {
        table[0][j] = b.slice(0,j).map(function(e,j){return {action:"insert",element:e,position:j}});
    }

    for(var i = 0; i < a.length; i++) {
        table[i+1] = [];
        table[i+1][0] = a.slice(0,i+1).map(function(e,i){return {action:"insert",element:e,position:i}});
        

        for(var j = 0; j < b.length; j++) {
            var f = [{action:"replace",position:j,element:[a[i],b[j]]}];
            if(a[i] === b[j]) {
                f = [];
            }
            console.log(table[i][j].concat(f),table[i+1][j],table[i][j+1])
            table[i+1][j+1] = [table[i][j].concat(f),table[i+1][j].concat([{action:"insert",position:j,element:b[j]}]),table[i][j+1].concat([{action:"delete",position:j,element:a[i+1]}])].sort(function(x,y){
                return x.length - y.length;
            })[0]

            //document.write(table[i+1][j+1].length);
            //document.write("\t");

        }
        //document.write("<br/>");
    }
    
    for(var i = 0; i <= a.length; i++) {

        for(var j = 0; j <= b.length; j++) {
            //if(i)
            document.write(table[i][j] && table[i][j].length);
            document.write("\t");
        } 
        document.write("<br/>");
    }
    document.write(table[a.length][b.length].map(function(e){return e.action + " " + e.element }).join("\n"));
}


//levenshtein("sailn".split(""),"failing".split(""));
levenshtein("ac".split(""),"a2c".split(""));
</script>
</pre>

感谢计子，又学到新东西了

话说计子，我觉得这个实现不是很高效呢？比如说用到拼写检查（editor常见需求），字典里每个词条都这么比一下，会占很多空间吧？