题目描述：
A strobogrammatic number is a number that looks the same when rotated 180 degrees (looked at upside down).
Write a function to count the total strobogrammatic numbers that exist in the range of low <= num <= high.
Example 1:
Input: low = “50”, high = “100”
Output: 3
Explanation: 69, 88, and 96 are three strobogrammatic numbers.
Note:
Because the range might be a large number, the low and high numbers are represented as string.
Time complexity: O(n)
沿用Leetcode 0247的思路，给出字符串长度n，我们可以通过递归找到长度为n的所有对称数。本题中我们枚举从low的长度到high的长度。在每次枚举中递归找到在该长度下的对称数 并且 满足 大于 low的值小于 high的值。以为本题要求只返回个数所以递归结果只需设为 int 。
class Solution {
private static final char[][] PAIRS = new char[][] {
{
'0', '0'}, {
'1', '1'}, {
'6', '9'}, {
'8', '8'}, {
'9', '6'}};
public int strobogrammaticInRange(String low, String high) {
if(low == null || high == null || low.length() > high.length()
|| (low.length() == high.length() && low.compareTo(high) > 0))
return 0;
int count = 0;
for (int len = low.length(); len <= high.length(); len++) {
count += dfs(low, high, new char[len], 0, len - 1);
}
return count;
}
private int dfs(String low, String high, char[] ch, int left, int right) {
if (left > right) {
String s = new String(ch);
if ((ch.length == low.length() && s.compareTo(low) < 0)
|| (ch.length == high.length() && s.compareTo(high) > 0)) {
return 0;
} else {
return 1;
}
}
int count = 0;
for(char[] p : PAIRS){
ch[left] = p[0];
ch[right] = p[1];
if (ch.length != 1 && ch[0] == '0') {
continue;
}
if (left == right && (p[0] == '6' || p[0] == '9')) {
continue;
}
count += dfs(low, high, ch, left + 1, right - 1);
}
return count;
}
}