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| 1 | +# 205. Isomorphic Strings |
| 2 | + |
| 3 | +- Difficulty: Easy. |
| 4 | +- Related Topics: Hash Table, String. |
| 5 | +- Similar Questions: Word Pattern. |
| 6 | + |
| 7 | +## Problem |
| 8 | + |
| 9 | +Given two strings `s` and `t`, **determine if they are isomorphic**. |
| 10 | + |
| 11 | +Two strings `s` and `t` are isomorphic if the characters in `s` can be replaced to get `t`. |
| 12 | + |
| 13 | +All occurrences of a character must be replaced with another character while preserving the order of characters. No two characters may map to the same character, but a character may map to itself. |
| 14 | + |
| 15 | + |
| 16 | +Example 1: |
| 17 | +``` |
| 18 | +Input: s = "egg", t = "add" |
| 19 | +Output: true |
| 20 | +```Example 2: |
| 21 | +``` |
| 22 | +Input: s = "foo", t = "bar" |
| 23 | +Output: false |
| 24 | +```Example 3: |
| 25 | +``` |
| 26 | +Input: s = "paper", t = "title" |
| 27 | +Output: true |
| 28 | +``` |
| 29 | + |
| 30 | +**Constraints:** |
| 31 | + |
| 32 | + |
| 33 | + |
| 34 | +- `1 <= s.length <= 5 * 104` |
| 35 | + |
| 36 | +- `t.length == s.length` |
| 37 | + |
| 38 | +- `s` and `t` consist of any valid ascii character. |
| 39 | + |
| 40 | + |
| 41 | + |
| 42 | +## Solution |
| 43 | + |
| 44 | +```javascript |
| 45 | +/** |
| 46 | + * @param {string} s |
| 47 | + * @param {string} t |
| 48 | + * @return {boolean} |
| 49 | + */ |
| 50 | +var isIsomorphic = function(s, t) { |
| 51 | + if (s.length !== t.length) return false; |
| 52 | + var map = {}; |
| 53 | + var used = {}; |
| 54 | + for (var i = 0; i < s.length; i++) { |
| 55 | + if (map[s[i]]) { |
| 56 | + if (map[s[i]] !== t[i]) return false; |
| 57 | + } else { |
| 58 | + if (used[t[i]]) return false; |
| 59 | + used[t[i]] = true; |
| 60 | + map[s[i]] = t[i]; |
| 61 | + } |
| 62 | + } |
| 63 | + return true; |
| 64 | +}; |
| 65 | +``` |
| 66 | + |
| 67 | +**Explain:** |
| 68 | + |
| 69 | +nope. |
| 70 | + |
| 71 | +**Complexity:** |
| 72 | + |
| 73 | +* Time complexity : O(n). |
| 74 | +* Space complexity : O(n). |
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