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| 1 | +想法:將所有字串擷取出來,存在 stack 裡面;接著再從頂部 pop 出並組成新字串即可 |
| 2 | + |
| 3 | +Time Complexity : O(n) for traversing string s |
| 4 | +Space Complexity : O(n) for the stack to store all strings (the strings total length <= s.length ) |
| 5 | + |
| 6 | +class Solution { |
| 7 | +public: |
| 8 | + string reverseWords(string s) { |
| 9 | + stack<string> store ; |
| 10 | + string temp ; |
| 11 | + for(auto &i : s) { |
| 12 | + if ( i == ' ' ) { |
| 13 | + if ( !temp.empty() ) { |
| 14 | + store.push(temp) ; |
| 15 | + temp = ""; |
| 16 | + } |
| 17 | + } |
| 18 | + else |
| 19 | + temp += i ; |
| 20 | + } |
| 21 | + if (!temp.empty()) |
| 22 | + store.push(temp); |
| 23 | + |
| 24 | + string ans ; |
| 25 | + while ( !store.empty() ) { |
| 26 | + if ( ans.empty() ) { |
| 27 | + ans += store.top() ; |
| 28 | + } |
| 29 | + else { |
| 30 | + ans += " " + store.top() ; |
| 31 | + } |
| 32 | + store.pop() ; |
| 33 | + } |
| 34 | + return ans ; |
| 35 | + } |
| 36 | +}; |
| 37 | + |
| 38 | +// 法二:如果不使用額外空間(除了輸出字串),就要把 stack 的過程直接作用在輸出字串上 |
| 39 | +也就是 ans = temp + " " + ans |
| 40 | + |
| 41 | +Time Complexity : O(n^2) for O(n) strings and each one take O(n) time to concate |
| 42 | +Space Complexity : O(n) for the answer string |
| 43 | + |
| 44 | + |
| 45 | +class Solution { |
| 46 | +public: |
| 47 | + string reverseWords(string s) { |
| 48 | + stack<string> store ; |
| 49 | + string temp ; |
| 50 | + for(auto &i : s) { |
| 51 | + if ( i == ' ' ) { |
| 52 | + if ( !temp.empty() ) { |
| 53 | + store.push(temp) ; |
| 54 | + temp = ""; |
| 55 | + } |
| 56 | + } |
| 57 | + else |
| 58 | + temp += i ; |
| 59 | + } |
| 60 | + if (!temp.empty()) |
| 61 | + store.push(temp); |
| 62 | + |
| 63 | + string ans ; |
| 64 | + while ( !store.empty() ) { |
| 65 | + if ( ans.empty() ) { |
| 66 | + ans += store.top() ; |
| 67 | + } |
| 68 | + else { |
| 69 | + ans += " " + store.top() ; |
| 70 | + } |
| 71 | + store.pop() ; |
| 72 | + } |
| 73 | + return ans ; |
| 74 | + } |
| 75 | +}; |
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