Question

当我输入这个函数85时，它只返回85.我不明白为什么不再以5为第一个数字递归调用自身。

console.log(PermutationStep(85));

function PermutationStep(num) { 
  var perms = [];

  rec(String(num), String(num).length, [], '');
  return perms;

  function rec(num, numLength, used, currPerm) {
    console.log(currPerm);
    if (currPerm.length === numLength) {
      perms.push(num);
    }
    for (var j=0; j<numLength; j++) {
      if (used[j]) continue;
      else {
        used[j]=true;
        rec(num, numLength, used, currPerm+num[j]);
      }
    }
  }
}

Answer 1

当达到篇幅时，您应该推送currPerm而不是num。
跟踪后再次使used[j]可用。
可能会产生重复的结果，例如885，您需要确保其不会跟踪相同的值。

console.log(PermutationStep(85));
console.log(PermutationStep(885));

function PermutationStep(num) { 
  var perms = [];
  
  // Convert number to char array.
  var str = ('' + num).split('');
  var length = str.length;
  // sort by alphabetical order.
  str.sort(function(a, b) {
    return a - b;
  });

  // Create used map.
  var used = str.map(function() {
    return false;
  });

  rec(str, '');
  return perms;

  // Not that javascript can reference value from outer function scope,
  // So we don't need to pass length and used each time, like you treat with perms.
  function rec(num, currPerm) {
    console.log(currPerm);
    if (currPerm.length === length) {
      // Put the constructed string here, not num
      perms.push(currPerm);
    }
    var prev = null, ch;
    for (var j = 0; j < length; j++) {
      if (used[j]) {
        continue;
      }
      ch = str[j];
      if (prev === ch) { 
        // If current char is same wih previous, there's more than 2 ch in the string,
        // So we need to skip to prevent duplicate result.
        continue;
      }
      used[j] = true;
      rec(num, currPerm + ch);
      // Make the number usable again.
      used[j]=false;
      prev = ch;
    }
  }
}

Javascript递归数字排列

1 个答案: