7. 整数反转
package 字符串; public class 整数反转 { public static void main(String[] args) { 整数反转 o = new 整数反转(); System.out.println(o.reverse2(-2147483412)); // System.out.println(Integer.MAX_VALUE);//2147483647 // System.out.println(Integer.MIN_VALUE);//-2147483648 } // 常规方法,先把数字转成字符串,然后反转,反转后再校验数字的合法性 public int reverse(int x) { int result = 0; boolean flag = false; if (x < 0) { x = -x; flag = true; } String s = String.valueOf(x); char[] c = s.toCharArray(); int i = 0; int j = c.length - 1; while (i < j) { char temp = c[i]; c[i] = c[j]; c[j] = temp; i++; j--; } // 翻转后的数字 String r = String.valueOf(c); // 判断是否有效 if ((r.length() == 10 && !isEffective(flag, r)) || r.length() > 10) { return 0; } result = Integer.parseInt(r); if (flag) { result = -result; } return result; } // flag标志正负数,ture是负数 public boolean isEffective(boolean flag, String r) { String compare = null; if (flag) { // compare = "2147483648"; String sMin = String.valueOf(Integer.MIN_VALUE); compare = sMin.substring(1, sMin.length()); } else { compare = String.valueOf(Integer.MAX_VALUE); // compare = "2147483647"; } // 2143847412 // 2147483648 for (int i = 0; i < r.length(); i++) { if (r.charAt(i) == compare.charAt(i)) { continue; } else if (r.charAt(i) > compare.charAt(i)) { return false; } else { return true; } } return false; } // 按位截取反转自己实现 public int reverse1(int x) { int result = 0; while (x != 0) { int k = x % 10; x = x / 10; if (result > Integer.MAX_VALUE / 10 || (result == Integer.MAX_VALUE / 10 && k > 7)) return 0; if (result < Integer.MIN_VALUE / 10 || (result == Integer.MIN_VALUE / 10 && k < -8)) return 0; result = result * 10 + k; } return result; } // 按位截取反转 public int reverse2(int x) { int rev = 0; while (x != 0) { //int的最大值是xxx7,最小值是-xxx8,模拟栈的操作,例如:8753,取第一个数,利用%,剩下的就是除以10取整 int k = x % 10; x = x / 10; //正数,当塞到剩最后一位时判断,发现大了肯定会溢出,等于的时候,最后一位不要大于7,如果溢出就返回0 //当获取的位数超过了最长位数-1,需要小心是否会溢出的问题了 if (rev > Integer.MAX_VALUE / 10 || (rev == Integer.MAX_VALUE && k > 7)) return 0; //负数 if (rev < Integer.MIN_VALUE / 10 || (rev == Integer.MIN_VALUE / 10 && k < -8)) return 0; //将该数追加到rev后端 rev = rev * 10 + k; } return rev; } }
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class Solution { public int reverse(int x) { // 121939423942394 int rev = 0; while (x != 0) { if (rev < Integer.MIN_VALUE/10 || rev > Integer.MAX_VALUE/10) { return 0; } else { rev = rev * 10 + x % 10; x = x / 10; } } return rev; } }