动态规划-数位dp-233. 数字 1 的个数

    int[] digits = new int[64];
    int[][] dp = new int[64][64];
    
    public int countDigitOne(int n) {
        if (n <= 0) return 0;  
        int pos = 0;
        while (n > 0) {
            digits[pos++] = n % 10;
            n /= 10;
        }
        return dfs(pos - 1, 0, true);
    }
    
    public int dfs(int pos, int cnt, boolean limit) {
        if (pos == -1) return cnt;
        if (!limit && dp[pos][cnt] != 0) return dp[pos][cnt];
        int up = limit ? digits[pos] : 9;
        int res = 0;
        for (int i = 0; i <= up; i++) {
            res += dfs(pos - 1, cnt + (i == 1 ? 1 : 0), limit && i == up);
        }
        if (!limit) dp[pos][cnt] = res;
        return res;
    }