LeetCode Js-205. Isomorphic Strings

Given two strings s and t, determine if they are isomorphic.
Two strings s and t are isomorphic if the characters in s can be replaced to get t.
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.

給予兩個字串 s 和 t，確認他們是否同構。
如果可以交換 s 和 t，則兩個字串為同構。
當保持字符順序時，必須替換所有出現的字符。
兩個字符不可對應到一個字符，但是可以對應到自己。

Example 1:

Input: s = "egg", t = "add"
Output: true

Example 2:

Input: s = "foo", t = "bar"
Output: false

Example 3:

Input: s = "paper", t = "title"
Output: true

Solution:
1. 先設定 objectS = {}, objectT = {}。
2. 運用 for 迴圈次數為 s 的長度，每一輪比對內容。
3. 如不同則返回 false
4. 其餘狀況將該輪的 object內容更新。
5. 完整走完迴圈則返回 ture。

Code 1:

var isIsomorphic = function(s, t) {
    if (s.length !== t.length) return false;

    let objectS = {}, objectT = {}
    
    for (let i = 0; i < s.length; i++) {
        if (objectS[s[i]] !== objectT[t[i]]) {
            return false
        } else {
            objectS[s[i]] = i
            objectT[t[i]] = i
        }
    }
    return true
}

FlowChart:
Example 1

Input: s = "egg", t = "add"

step.1
i = 0
objectS[s[0]] = underlined, objectT[t[0]] = underlined //===
objectS[s[0]] = i //0
objectT[t[0]] = i //0

step.2
i = 1 
objectS[s[1]] = underlined, objectT[t[1]] = underlined //===
objectS[s[1]] = i //1
objectT[t[1]] = i //1

step.3
i = 2 
objectS[s[2]] = underlined, objectT[t[2]] = underlined //===
objectS[s[2]] = i //2
objectT[t[2]] = i //2

return true

Example 2

Input: s = "foo", t = "bar"

step.1
i = 0
objectS[s[0]] = underlined, objectT[t[0]] = underlined //===
objectS[s[0]] = i //0
objectT[t[0]] = i //0

step.2
i = 1 
objectS[s[1]] = underlined, objectT[t[1]] = underlined //===
objectS[s[1]] = i //1
objectT[t[1]] = i //1

step.3
i = 2 
objectS[s[2]] = 1, objectT[t[2]] = underlined //!==
return false

Example 3

Input: s = "paper", t = "title"

step.1
i = 0
objectS[s[0]] = underlined, objectT[t[0]] = underlined //===
objectS[s[0]] = i //0
objectT[t[0]] = i //0

step.2
i = 1 
objectS[s[1]] = underlined, objectT[t[1]] = underlined //===
objectS[s[1]] = i //1
objectT[t[1]] = i //1

step.3
i = 2 
objectS[s[2]] = 0, objectT[t[2]] = 0 //===
objectS[s[2]] = i //2
objectT[t[2]] = i //2

step.4
i = 3
objectS[s[3]] = underlined, objectT[t[3]] = underlined //===
objectS[s[3]] = i //3
objectT[t[3]] = i //3

step.5
i = 4
objectS[s[4]] = underlined, objectT[t[4]] = underlined //===
objectS[s[4]] = i //4
objectT[t[4]] = i //4

step.6
i = 5 
objectS[s[5]] = underlined, objectT[t[5]] = underlined //===
objectS[s[5]] = i //5
objectT[t[5]] = i //5

return true

Code 2:

var isIsomorphic = function(s, t) {
    if (s.length !== t.length) return false;

    for (let i = 0; i < s.length; i++) {
        if (s.indexOf(s[i]) !== t.indexOf(t[i])) {
            return false;
        }
    }
    return true;
};