解决河内塔问题。
您的解决方案应该接受光盘数量作为第一个参数,并使用三个字符串来识别三个光盘堆栈中的每一个,例如towerOfHanoi(4, 'A', 'B', 'C')
。该函数应该返回一个包含移动列表的数组数组,source - > destination。例如,数组[['A', 'C'], ['B', 'A']]
表示第一个移动是将光盘从堆栈A移动到C,第二个移动是移动一个从堆栈B到A的光盘
towerOfHanoi
是一个功能。
testString: 'assert(typeof towerOfHanoi === "function", "towerOfHanoi
is a function.");'
- text: ''
testString: 'assert(res3.length === 7, "towerOfHanoi(3, ...)
should return 7 moves.");'
- text: 'towerOfHanoi(3, "A", "B", "C")
应返回[[“A”,“B”],[“A”,“C”],[“B”,“C”],[ “A”, “B”],[ “C”, “A”],[ “C”, “B”],[ “A”, “B”]]“)。'
testString: 'assert.deepEqual(towerOfHanoi(3, "A", "B", "C"), res3Moves, "towerOfHanoi(3, "A", "B", "C")
should return [["A","B"],["A","C"],["B","C"],["A","B"],["C","A"],["C","B"],["A","B"]].");'
- text: 'towerOfHanoi(5, "X", "Y", "Z")
第10 towerOfHanoi(5, "X", "Y", "Z")
应为Y - > X.'
testString: 'assert.deepEqual(res5[9], ["Y", "X"], "towerOfHanoi(5, "X", "Y", "Z")
10th move should be Y -> X.");'
- text: 'towerOfHanoi(7, "A", "B", "C")
前十个动作是[[“A”,“B”],[“A”,“C”],[“B”,“C”] [ “A”, “B”],[ “C”, “A”],[ “C”, “B”],[ “A”, “B”],[ “A”, “C”] [ “B”, “C”],[ “B”, “A”]]“)。'
testString: 'assert.deepEqual(towerOfHanoi(7, "A", "B", "C").slice(0, 10), res7First10Moves, "towerOfHanoi(7, "A", "B", "C")
first ten moves are [["A","B"],["A","C"],["B","C"],["A","B"],["C","A"],["C","B"],["A","B"],["A","C"],["B","C"],["B","A"]].");'
```