博文中会简要介绍Leetcode P0246题目分析及解题思路。
“Strobogrammatic Number”是一道比较简单的题目，主要思路是建立数字和其旋转180度以后的数字之间的映射（如果旋转后能够构成一个数字的话），然后以数字为单位遍历一遍当前的数即可。
A strobogrammatic number is a number that looks the same when rotated 180 degrees (looked at upside down).
Write a function to determine if a number is strobogrammatic. The number is represented as a string.
Example 1:
Input: num = "69" Output: true
以下是Java的题解代码实现。
class Solution {
private Map<Character, Character> ch2CharMap = null;
public Solution() {
this.ch2CharMap = new HashMap<>();
this.ch2CharMap.put('6', '9');
this.ch2CharMap.put('9', '6');
this.ch2CharMap.put('1', '1');
this.ch2CharMap.put('0', '0');
this.ch2CharMap.put('8', '8');
}
public boolean isStrobogrammatic(String num) {
int L = num.length();
for (int idx=0; idx<(L+1)/2; ++idx) {
char ch = num.charAt(idx);
if (this.ch2CharMap.containsKey(ch)) {
char mapped = this.ch2CharMap.get(ch);
if (mapped != num.charAt(L-1-idx))
return false;
}
else
return false;
}
return true;
}
}
以下是C++的题解代码实现。
class Solution {
public:
Solution() {
ch2ch_['6'] = '9';
ch2ch_['9'] = '6';
ch2ch_['8'] = '8';
ch2ch_['1'] = '1';
ch2ch_['0'] = '0';
}
bool isStrobogrammatic(string num) {
int L = num.length();
for (int idx=0; idx<(L+1)/2; ++idx) {
if (ch2ch_.count(num[idx])) {
char mapped = num[L-1-idx];
if (mapped != ch2ch_[num[idx]])
return false;
}
else
return false;
}
return true;
}
private:
unordered_map<char, char> ch2ch_;
};