題目描述:
代碼實現:
- 思路很簡單,關鍵在於對於邊界值的處理
- 時間複雜度:O(log(x))
/**
* @param {number} x
* @return {number}
*/
var reverse = function(x) {
var rev = 0
while (x !== 0) {
var pop = x % 10
x = parseInt(x / 10)
if (rev > 2147483647 / 10 || (rev == 2147483647 / 10 && pop > 7)) return 0
if (rev < -2147483648 / 10 || (rev == -2147483648 / 10 && pop < -8)) return 0
rev = rev * 10 + pop
}
return rev
};