使括号平衡的最小交换次数
给定 2N 长度的字符串，包含N个 '[' ，N个 ']'，计算使字符串平衡的最小交换次数
平衡字符串定义：S1[S2]， 其中 S1，S2均为平衡字符串
2N长度字符串
最小交换次数
Input : []][][Output : 2First swap: Position 3 and 4 [][]][Second swap: Position 5 and 6 [][][]Input : [[][]]Output : 0String is already balanced.
Solution is below
顺序遍历字符串，当遇到不匹配的']'时（遇到相应的'['之前），将其与之后最接近它的']'交换位置
Python
C++
def MinSwapBracketBalancing(Brackets: list):
# 首先遍历字符串，获取所有左括号的位置
leftBrackets = []
leftBrackets.extend([ idx for idx, x in enumerate(Brackets) if x == '['])
if len(leftBrackets) != len(Brackets) // 2:
return 0
count = 0 # 当前还未平衡的左括号数量
pos = 0 # 下一个左括号的位置
sum = 0 # 总共移动次数
i = 0
while i < len(Brackets):
if Brackets[i] == '[':
# 当前遍历括号为左括号
count += 1
pos += 1
else:
# 当前括号为右括号
count -= 1 # 不平衡左括号 - 1
if count < 0:
# 右括号在不平衡左括号之前出现
# 交换两括号位置
Brackets[i], Brackets[leftBrackets[pos]] = Brackets[leftBrackets[pos]],Brackets[i]
sum += leftBrackets[pos] - i
count = 0
pos += 1
i += 1
i += 1
return sum
int MinSwapsBracketBalancing(string Brackets) {
vector<int> leftBrackets;
// Get positions of all left brackets
for (auto i = 0;i<Brackets.size(); i++) {
if (Brackets[i] == '[')
leftBrackets.push_back(i);
}
auto count = 0; // Current unbalanced left brackets
auto pos = 0; // Next position of next left brackets
auto sum = 0; // Total moves to balance brackets
for (auto i = 0; i < Brackets.size(); i++) {
if (Brackets[i] == '[') {
count++;
pos++;
}
else if (--count < 0) {
// Get unbalanced right bracket
// Swap it with next left bracket
swap(Brackets[i], Brackets[leftBrackets[pos]]);
sum += leftBrackets[pos] - i;
}
}
return sum;
}
Last modified 4yr ago