leetcode 第74场周赛

https://leetcode.com/contest/weekly-contest-74

1、Valid Tic-Tac-Toe State

井字棋，判断在下的过程中，会不会出现题目给的局面。

import java.util.Arrays;

/**
 * Created by dezhonger on 2019/7/30
 */
public class Leetcode0794 {
    public boolean validTicTacToe(String[] board) {
        int a, b;
        char c1, c2;
        a = b = 0;
        c1 = 'X';
        c2 = 'O';
        //分别表示两个人形成横着的，竖着的，斜对角三连个数
        int[] gx = new int[3];
        int[] go = new int[3];
        for (String s : board) {
            if (s.equals("XXX")) gx[0]++;
            if (s.equals("OOO")) go[0]++;
            for (char c : s.toCharArray()) {
                if (c == c1) a++;
                if (c == c2) b++;
            }
        }
        if (b > a) return false;
        if (a - b > 1) return false;

        for (int i = 0; i < 3; i++) {
            if (board[0].charAt(i) == c1 && board[1].charAt(i) == c1 && board[2].charAt(i) == c1) gx[1]++;
            if (board[0].charAt(i) == c2 && board[1].charAt(i) == c2 && board[2].charAt(i) == c2) go[1]++;
        }
        if (board[0].charAt(0) == c1 && board[1].charAt(1) == c1 && board[2].charAt(2) == c1) gx[2]++;
        if (board[0].charAt(0) == c2 && board[1].charAt(1) == c2 && board[2].charAt(2) == c2) go[2]++;
        if (board[0].charAt(2) == c1 && board[1].charAt(1) == c1 && board[2].charAt(0) == c1) gx[2]++;
        if (board[0].charAt(2) == c2 && board[1].charAt(1) == c2 && board[2].charAt(0) == c2) go[2]++;
        //g1, g2分别表示两个人形成三连的个数
        int g1, g2;
        g1 = gx[0] + gx[1] + gx[2];
        g2 = go[0] + go[1] + go[2];
        //不可能俩人同时赢
        if (g1 >= 1 && g2 >= 1) return false;
        //不可能出现横着或竖着的有两个
        if (gx[0] > 1 || gx[1] > 1) return false;
        if (go[0] > 1 || go[1] > 1) return false;
        //a赢的话，棋子个数必须必b多一个
        if (g1 > 0 && a != b + 1) return false;
        //b赢的话，旗子个数必须相等
        if (g2 > 0 && a != b) return false;
        return true;
    }
}

2、Number of Matching Subsequences

判断一些字符串是不是原字符串的子序列

做法：dp的思想，先预处理出每个位置后a到z下一次出现的位置，然后用这个进行转移

class Solution {

    //contest74
    public int numMatchingSubseq(String S, String[] words) {
        char[] s = S.toCharArray();
        int[][] next = getNext(s);
        int ret = 0;
        for (String w : words) {
            int pos = 0;
            int add = 1;
            for (char c : w.toCharArray()) {
                pos = next[pos][c - 'a'];
                if (pos > s.length) {
                    add = 0;
                    break;
                }
            }
            ret += add;
        }
        return ret;
    }

    public int[][] getNext(char[] s) {
        int n = s.length;
        //next[i][j] 第i个字幕后字母j出现的位置, 1-base
        int[][] next = new int[n + 1][26];
        //第n个字母是结尾了，没有字母了
        for (int i = 0; i < 26; i++) next[n][i] = n + 1;
        for (int i = s.length - 1; i >= 0; i--) {
            for (int j = 0; j < 26; j++) next[i][j] = next[i+1][j];
            next[i][s[i] - 'a'] = i + 1;
        }
        return next;
    }
}

3、Number of Subarrays with Bounded Maximum

求满足这样的子数组的个数，子数组的最大值在[L, R]之间

class Solution {
    //计算最大值<=r的子数组的个数
    private int f(int[] A, int r) {
        int c = 0;
        int res = 0;
        for (int x : A) {
            //c表示以x结尾的子数组个数，满足最大值<=r
            if (x <= r) c++;
            else c = 0;
            res += c;
        }
        return res;
    }
    
    public int numSubarrayBoundedMax(int[] A, int L, int R) {
        //最大值<=L的子数组，是最大值<=R的子数组的子集，相减就是最大值在[L, R]的子数组
        return f(A, R) - f(A, L - 1);
    }
}

4、Preimage Size of Factorial Zeroes Function

求阶乘最后0的个数是k的数字个数

注意到答案要么是0，要么是5，二分一下就可以了

public class Leetcode0793 {
    public static void main(String[] args) {
        new Leetcode0793().preimageSizeFZF(79);
    }
    //cal zeros of 5*x
    int f(int x) {
        int res = x;
        while (x > 0) {
            res += x / 5;
            x /= 5;
        }
        return res;
    }

    public int preimageSizeFZF(int K) {
        int r = K;
        int l = 0;
        while (l < r) {
            int mid = (l + r) / 2;
            int cnt = f(mid);
            if (cnt == K) return 5;
            if (cnt < K) l = mid + 1;
            else r = mid - 1;
        }
        return f(l) == K ? 5 : 0;
    }
}