[抄题]:

Given two strings s and t, determine if they are isomorphic.

Two strings 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.

For example,
Given "egg""add", return true.

Given "foo""bar", return false.

Given "paper""title", return true.

Note:
You may assume both s and t have the same length.

[暴力解法]:

时间分析:

空间分析:

[优化后]:

时间分析:

空间分析:

[奇葩输出条件]:

[奇葩corner case]:

[思维问题]:

[一句话思路]:

“对应类问题”:一批标记一次,标记对不上的不行

[输入量]:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入):

[画图]:

[一刷]:

[二刷]:

[三刷]:

[四刷]:

[五刷]:

[五分钟肉眼debug的结果]:

[总结]:

对应类也可以用256,一批标记一次, 标记都是用index i

[复杂度]:Time complexity: O(n) Space complexity: O(n)

[英文数据结构或算法,为什么不用别的数据结构或算法]:

不知道怎么用哈希,但其实256就是哈希的一种。

[关键模板化代码]:

一批标记一次

for (int i = 0; i < n; i++) {
if (m1[s.charAt(i)] != m2[t.charAt(i)]) return false;
m1[s.charAt(i)] = i + 1;
m2[t.charAt(i)] = i + 1;
}

[其他解法]:

[Follow Up]:

[LC给出的题目变变变]:

290. Word Pattern

[代码风格] :

class Solution {
public boolean isIsomorphic(String s, String t) {
//ini, int[256]
int[] m1 = new int[256];
int[] m2 = new int[256];
int n = s.length(); //judge
for (int i = 0; i < n; i++) {
if (m1[s.charAt(i)] != m2[t.charAt(i)]) return false;
m1[s.charAt(i)] = i + 1;
m2[t.charAt(i)] = i + 1;
} //return false;
return true;
}
}

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