Question

我正在尝试编写一个函数来执行其中的注释中清楚描述的内容：

function single_ops(w)
    {
        /* Given a word w, return an array of all strings that are w with:

           (1) 1 letter inserted, that letter being one which is distance
               1 away on the keyboard
           (2) 1 letter replaced, that letter being one which is distance
               1 away on the keyboard
           (3) 1 letter removed
           (4) 2 adjacent letters swapped
        */

    }

我写了一个有效的程序，但它非常混乱！

        var A = new Array(); // Associative array to store words, so we can add them w/out addding repeats
        for (var i = 0, j = w.length; i < j; ++i)
        {
            var closeChars = kdmap[w[i]];
            var s1a = w.slice(0,i); // equals w[0], ..., w[i-1]    
            var s2a = w.slice(i,j); // equals w[i], ..., w[j-1]    
            var s1b = w.slice(0,i+1); // equals w[0], ..., w[i]
            var s2b = w.slice(i+1,j) // equals w[i+1], ..., w[j-1]
            for (var k = 0, n = closeChars.length; k < n; ++k)
            {
                var cc = closeChars[k];
                // (1) ----------------------------------------------
                A[s1a + cc + s2a] = 1; // equals w[0],..., w[i-1], cc, w[i],..., w[j-1]
                A[s1b + cc + s2b] = 1; // equals w[0], ..., w[i], cc, w[i+1], ..., w[j-1]
                // (2) ----------------------------------------------
                A[s1a + cc + s2b] = 1; // equals w[0], ..., w[i-1], cc, w[i+1], ..., w[j-1]
            }
            // (3) --------------------------------------------------
            if (j > 1)
                A[s1a + s2b] = 1; // equals w[0], ..., w[i-1], w[i+1], ..., w[j-1]
            // (4) --------------------------------------------------
            A[s1a + s2b.slice(0,1) + s2a.slice(0,1) + s2b.slice(1,s2b.length)] = 1;
        }
        return A;

我只是向你展示我的工作代码，所以大家不认为我是想让你为我写的。 ;）

我有几个问题：

字符串添加（就像我有）是在JavaScript中组合字符串的最有效方法还是有另一种方法？
我有if (j > 1)，因为我不希望数组中有任何空字符串。显然，如果条件总是以相同的方式进行评估，那么在循环中设置条件并不理想，但是我的替代方案是什么？
JavaScript关联数组是如何实际实现的？我正在使用我的，好像它是一个哈希表，但我不确定它是否在浏览器解释我的JavaScript时以这种方式实现。

哦，这是kdmap，我做的“关键距离图”：

   /* Map from each character on the keyboard to the characters that 
      are of distance 1 from it:
   */
    var kdmap = {'a': ['q','w','s','z'],
                 'b': ['v','g','h','n'],
                 'c': ['x','d','f','v'],
                 'd': ['s','e','r','f','c','x'],
                 'e': ['w','s','d','r'],
                 'f': ['d','r','t','g','v','c'],
                 'g': ['f','t','y','h','b','v'],
                 'h': ['g','y','u','j','n','b'],
                 'i': ['u','h','j','i'],
                 'j': ['h','n','m','k','i','u','y'],
                 'k': ['j','i','o','l','m'],
                 'l': ['k','o','p'],
                 'm': ['n','j','k'],
                 'n': ['b','h','j','m'],
                 'o': ['i','k','l','p'],
                 'p': ['o','l'],
                 'q': ['a','w'],
                 'r': ['e','d','f','t'],
                 's': ['a','w','e','d','x','z'],
                 't': ['r','f','g','y'],
                 'u': ['y','h','j','i'],
                 'v': ['c','f','g','b'],
                 'w': ['q','a','s','e'],
                 'x': ['z','s','d','c'],
                 'y': ['t','g','h','u'],
                 'z': ['a','s','x']
                };

JavaScript配方获取字符串S并返回S的所有1操作修改

0 个答案: