LeetCode 248. Strobogrammatic Number III

原题网址：https://leetcode.com/problems/strobogrammatic-number-iii/

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.

For example,
Given low = "50", high = "100", return 3. Because 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.

思路：深度优先搜索。注意当low的位数和high的位数不相等的时候如何处理，以及如何保证深度优先的生成的数字符合要求。

这是一个典型的深度优先搜索题目，我想在字符数值比较等地方提升搜索效率，于是写得比较恶心。

public class Solution {    private char[] pair = {'0','1','6','8','9'};    private char[] rotated = {'0', '1', ' ', ' ', ' ', ' ', '9', ' ', '8', '6'};    private char[] self = {'0','1','8'};    private int find(char[] low, char[] high, boolean min, boolean max, char[] buf, int step) {        if (step == buf.length) return 1;        int count = 0;        if ((step<<1) == buf.length) {            // 当位数为偶数，且深度优先过半时            for(int i=step, j=step-1; i<buf.length; i++, j--) {                buf[i] = rotated[buf[j]-'0'];                if (min && buf[i] < low[i]) return 0;                if (max && buf[i] > high[i]) return 0;                if (buf[i] > low[i]) min = false;                if (buf[i] < high[i]) max = false;            }            // System.out.printf("found %s\n", new String(buf));            return 1;        } else if ((step<<1) == buf.length-1) {            // 当位数为奇数，且深度优先处于中间数位时            for(int i=0; i<self.length; i++) {                boolean nmin = min;                boolean nmax = max;                buf[step] = self[i];                if (min && buf[step] < low[step]) continue;                if (max && buf[step] > high[step]) continue;                if (buf[step] > low[step]) nmin = false;                if (buf[step] < high[step]) nmax = false;                count += find(low, high, nmin, nmax, buf, step+1);            }        } else if ((step<<1) > buf.length) {            // 当位数为奇数，且深度优先过半时            for(int i=step, j=step-2; i<buf.length; i++, j--) {                buf[i] = rotated[buf[j]-'0'];                if (min && buf[i] < low[i]) return 0;                if (max && buf[i] > high[i]) return 0;                if (buf[i] > low[i]) min = false;                if (buf[i] < high[i]) max = false;            }            // System.out.printf("found %s\n", new String(buf));            return 1;        } else {            // 深度优先未过半时            for(int i=0; i<pair.length; i++) {                boolean nmin = min;                boolean nmax = max;                if (min && pair[i] < low[step]) continue;                if (max && pair[i] > high[step]) continue;                if (pair[i] > low[step]) nmin = false;                if (pair[i] < high[step]) nmax = false;                buf[step] = pair[i];                count += find(low, high, nmin, nmax, buf, step+1);            }        }        return count;    }    public int strobogrammaticInRange(String low, String high) {        if (low.length() > high.length()) return 0;        char[] ha = high.toCharArray();        char[] la = low.toCharArray();        int lw = la.length;        int count = 0;        do {            lw = la.length;            char[] upper = ha;            if (la.length < upper.length) {                upper = new char[la.length];                Arrays.fill(upper, '9');            }            // System.out.printf("low=%s, high=%s\n", new String(la), new String(upper));            count += find(la, upper, true, true, new char[la.length], 0);            if (lw < ha.length) {                la = new char[lw+1];                Arrays.fill(la, '0');                la[0] = '1';            }        } while (lw < ha.length);        return count;    }}

方法二：当涉及到区间[A,B]的计算时，有个重要的思路是将区间转换为[0,B]-[0,A)

另外，深度优先不一定只是深度加1，可以是从两边向中间搜索。

public class Solution {    private char[][] strobos = {{'0','0'}, {'1','1'}, {'6','9'}, {'8','8'}, {'9','6'}};    public int strobogrammaticInRange(String low, String high) {        if (low.length() > high.length()) return 0;        if (low.length() == high.length() && low.compareTo(high) > 0) return 0;        return count(high, true) - count(low, false);    }    private int count(String high, boolean inclusive) {        int len = high.length();        int count = 0;        for(int i=1; i<len; i++) {            count += countLen(i, true);        }        count += find(new char[len], 0, len-1, high.toCharArray(), inclusive, true);        return count;    }    private int countLen(int len, boolean outside) {        if (len == 0) return 1;        if (len == 1) return 3;        if (outside) return countLen(len-2, false)*4;        return countLen(len-2, false)*5;    }        private int find(char[] num, int left, int right, char[] high, boolean inclusive, boolean ceiling) {        if (left > right) {            if (!ceiling) return 1;            int compare = compare(num, high, left, num.length-1);            if (inclusive && compare <= 0) return 1;            if (!inclusive && compare < 0) return 1;            return 0;        }        int count = 0;        for(int i=0; i<strobos.length; i++) {            if (left == 0 && right != 0 && strobos[i][0] == '0') continue;            if (left == right && strobos[i][0] != strobos[i][1]) continue;            if (ceiling && strobos[i][0] > high[left]) continue;            num[left] = strobos[i][0];            num[right] = strobos[i][1];            count += find(num, left + 1, right - 1, high, inclusive, ceiling && num[left] == high[left]);        }        return count;    }    private int compare(char[] a, char[] b, int from, int to) {        for(int i=from; i<=to; i++) {            if (a[i] < b[i]) return -1;            if (a[i] > b[i]) return 1;        }        return 0;    }}

参考文章：

https://leetcode.com/discuss/73721/easiest-20ms-94%25-java-solution

https://leetcode.com/discuss/91136/java-0ms-solution-99-5%25