Question

var rockPaperScissors = function(rounds) {
  var outcomes = [];
  var plays = ['rock', 'paper', 'scissors'];
  var playedSoFar = [];

  var combos = function(roundsToGo) {
    // base case
    if (roundsToGo === 0) {
      outcomes.push(playedSoFar.slice());
      return;
    }

    for (var i = 0; i < plays.length; i++) {
      playedSoFar.push(plays[i]);
      combos(roundsToGo - 1);
      playedSoFar.pop();
    }
  };
  combos(rounds);
  return outcomes;
};

console.log(rockPaperScissors(2));

如果我从slice()取出playedSoFar，outcomes只会返回9个空数组，而不是2轮石头剪刀游戏中的所有组合：

[ [ 'rock', 'rock' ],
  [ 'rock', 'paper' ],
  [ 'rock', 'scissors' ],
  [ 'paper', 'rock' ],
  [ 'paper', 'paper' ],
  [ 'paper', 'scissors' ],
  [ 'scissors', 'rock' ],
  [ 'scissors', 'paper' ],
  [ 'scissors', 'scissors' ] ]

当我从slice()中取出playedSoFar时，为什么这不起作用？

Answer 1

.slice()会返回您的数组的副本，如果您不使用.slice()并只推送playedSoFar数组，那么其更新的内容将反映在整个范围内，因此它们将会反映在combos数组中。更多详情here。

Answer 2

如果你想保留playingSoFar，那么这就是你要走的路：

  outcomes=outcomes.concat(playedSoFar);

岩纸剪刀递归

2 个答案: