解法一:遍历
思想:
遍历存到res中
复杂度:
●时间:O(N)
●空间:O(N)
代码:
class Solution {
public:
string reverseLeftWords(string s, int n) {
string res = "";
for(int i = n; i < s.size(); ++i)
res += s[i];
for(int i = 0; i < n; ++i)
res += s[i];
return res;
}
};
解法二:substr
思想:
复杂度:
class Solution {
public:
string reverseLeftWords(string s, int n) {
string right = s.substr(0, n);
string left = s.substr(n, s.size() - n);
left += right;
return left;
}
};
解法三:三次反转
思想:
操作是左闭右开的
begin()是开头元素,end()是末尾 + 1的元素
代码:
class Solution {
public:
string reverseLeftWords(string s, int n) {
reverse(s.begin(), s.begin() + n);
reverse(s.begin() + n, s.end());
reverse(s.begin(), s.end());
return s;
}
};
解法四:一行
class Solution {
public:
string reverseLeftWords(string s, int n) {
return (s + s).substr(n, s.size());
}
};