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2021/04/20 - 回文数 #76
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题目
思路一(字符串匹配)
将数字转成字符串
str
,定义两个指针一头一尾i
和j
,两个指针分别移动,假如str[i]
等于str[j]
相等,则说明这一对字符是匹配的,则i
增1
,j
减1
,继续循环,直到i
大于j
;如果不相等,说明不匹配,则跳出循环,返回false
;循环匹配完成没有不等则返回true
临界情况处理
0
的数,它必然不是回文数,则直接返回false
0
的数,它必然不是回文数,则直接返回false
true
代码实现
思路二(数字翻转)
通过获取
a
翻转过来的数字b
,来判断两者是否相等判断思路
x
是偶数,则根据翻转后的数字rn
与x
的前一半比较,如果相等,则说明是回文数x
是奇数,则根据翻转后的数字rn
退一位的值与x
的前一半比较,如果相等,则说明是回文数。即如12521
,此时x
为12
,rn
为125
,只需判断x === Math.floor(rn / 10)
即可代码实现
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