leetcode660 remove 9

思路1：

二分+数位dp。

实现：

class Solution 
{
public:
    int dfs(int now, bool ok, bool flag, int dp[][2], int num[])
    {
        if (now == 0) return ok;
        if (!flag && dp[now][ok] != -1)
            return dp[now][ok];
        int res = 0;
        int u = flag ? num[now] : 9;
        for (int i = 0; i <= u; i++)
        {
            res += dfs(now - 1, ok || (i == 9), flag && i == u, dp, num);
        }
        return flag ? res : dp[now][ok] = res;
    }
    int newInteger(int n) 
    {
        long long l = n, r = 5e9, res = n;
        int dp[15][2], num[15]; 
        while (l <= r)
        {
            long long m = (l + r) >> 1;
            memset(dp, -1, sizeof dp);
            memset(num, 0, sizeof num);
            int sum = 0;
            long long tmp = m;
            while (tmp)
            {
                num[++sum] = tmp % 10;
                tmp /= 10;
            }
            int ans = dfs(sum, 0, true, dp, num);
            if (m - ans < n) l = m + 1;
            else 
            {
                r = m - 1;
                if (m - ans == n)
                    res = m;
            }
        }
        return res;
    }
};

思路2：

去掉包含9的数之后实际上就是变成了9进制。

实现：

class Solution 
{
public:
    int newInteger(int n) 
    {
        int sum = 0, x = 1;
        while (n)
        {
            sum += n % 9 * x;
            n /= 9;
            x *= 10;
        }
        return sum;
    }
};