Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses.
For example, given n = 3, a solution set is:
[
"((()))",
"(()())",
"(())()",
"()(())",
"()()()"
]
Tags: String, Backtracking
题意是给你 n
值,让你找到所有格式正确的圆括号匹配组,题目中已经给出了 n = 3
的所有结果。遇到这种问题,第一直觉就是用到递归或者堆栈,我们选取递归来解决,也就是 helper
函数的功能,从参数上来看肯定很好理解了,leftRest
代表还有几个左括号可以用,rightNeed
代表还需要几个右括号才能匹配,初始状态当然是 rightNeed = 0, leftRest = n
,递归的终止状态就是 rightNeed == 0 && leftRest == 0
,也就是左右括号都已匹配完毕,然后把 str
加入到链表中即可。
func generateParenthesis(_ n: Int) -> [String] {
var list: [String] = [String]()
helper(&list, "", 0, n)
return list
}
func helper(_ list:inout [String], _ tempStr: String, _ rightNeed: Int, _ leftRest: Int) {
if rightNeed == 0 && leftRest == 0 {
list.append(tempStr)
return
}
if rightNeed > 0 {
helper(&list, tempStr + ")", rightNeed - 1, leftRest)
}
if leftRest > 0 {
helper(&list, tempStr + "(", rightNeed + 1, leftRest - 1)
}
}
另一种实现方式就是迭代的思想了,我们来找寻其规律如下所示:
f(0): “”
f(1): “(“f(0)”)”
f(2): "(“f(0)”)"f(1), “(“f(1)”)”
f(3): "(“f(0)”)"f(2), "(“f(1)”)"f(1), “(“f(2)”)”
...
可以递推出 f(n) = "(“f(0)”)"f(n-1) , "(“f(1)”)"f(n-2) "(“f(2)”)"f(n-3) … "(“f(i)”)“f(n-1-i) … “(f(n-1)”)”
根据如上递推式写出如下代码应该不难了吧。
func generateParenthesis(_ n: Int) -> [String] {
var queueListDic: [Int: [String]] = [Int: [String]]()
queueListDic[0] = [""]
for i in 1...n {
var list: [String] = [String]()
for j in 0..<i {
for itemJ in queueListDic[j]! {
for itemIMinusJ in queueListDic[i - j - 1]! {
list.append("(" + itemJ + ")" + itemIMinusJ)
}
}
}
queueListDic[i] = list
}
return queueListDic[n]!
}
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