<p>Given an integer <code>n</code>, return the <code>n<sup>th</sup></code> digit of the infinite integer sequence <code>[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ...]</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 3
<strong>Output:</strong> 3
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 11
<strong>Output:</strong> 0
<strong>Explanation:</strong> The 11<sup>th</sup> digit of the sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ... is a 0, which is part of the number 10.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 2<sup>31</sup> - 1</code></li>
</ul>

class Solution {
public:
    int findNthDigit(int n) {

        // 1~9 : 10 - 1
        // 10~99 : 100 - 10
        // 100~999: 1000 - 100
        int curDigit = 0;
        int curDigitNum = 1;
        long long nextAddDigit = 10;
        long long cntDigit = 0;
        while(cntDigit < n) {
            curDigit++;
            if(nextAddDigit == curDigit) {
                nextAddDigit *= 10;
                curDigitNum++;
            }
            cntDigit+=curDigitNum;
        }
        string num = to_string(curDigit);
        printf("%d %s\n", cntDigit, num.c_str());
        return num[num.size() - 1 - (cntDigit-n)] - '0';
    }
};
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