从数组javascript返回最小可能的联接 [英] return the smallest possible join from array javascript

问题描述

我有适合以下数组的代码:['45'，'30'，'50'，'1'].

I have this code which works good for this array: ['45', '30', '50', '1'].

function penalty(a_list) {
  return a_list.sort((a, b) => a - b).join('');
}

例如:假设a_list为['45'，'30'，'50'，'1']，则最小的字符串为'1304550'，这是正确的，但是如果a_list为['32'，给定当前代码，"3"]我将得到不正确的"332"，因为"323"是可能的最小字符串.希望能有所帮助.

for example: given that a_list is ['45', '30', '50', '1'] the smallest possible string will be '1304550' which is right, but what if a_list is ['32', '3'] given this current code I will get '332' which is not correct because '323' is the smallest possible string. Hope that helps.

推荐答案

您可以采用a和b的隐式值以及b和a的值，并取其差值进行排序，它以较小的值反映了两个字符串的排序顺序，以供以后连接.

You could take the concatinated values of a and b and the value of b and a and take the delta of it for sorting, which reflects the sort order of the two string for a smaller value for later joining.

如果提供了整数，则需要在sort回调中将值转换为字符串.

If integers are supplied, then the values need to be converted to string in the sort callback.

function sort(a, b) {
    return (a + b) - (b + a);
}

console.log([['45', '30', '50', '1'], ['32', '3']].map(a => a.sort(sort).join('')));

对于稳定的排序，可以将较小的值移到顶部(这不会影响后面的连接字符串).这将'3'排在'33'之前.

For a stable sort, you could move smaller values to top (which does not affect the later joined string). This sorts '3' before '33'.

function sort(a, b) {
    return (a + b) - (b + a) || a - b;
}

console.log([['45', '30', '50', '1'], ['32', '3']].map(a => a.sort(sort).join('')));

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