DI String Match
<p>A permutation <code>perm</code> of <code>n + 1</code> integers of all the integers in the range <code>[0, n]</code> can be represented as a string <code>s</code> of length <code>n</code> where:</p>

<ul>
	<li><code>s[i] == &#39;I&#39;</code> if <code>perm[i] &lt; perm[i + 1]</code>, and</li>
	<li><code>s[i] == &#39;D&#39;</code> if <code>perm[i] &gt; perm[i + 1]</code>.</li>
</ul>

<p>Given a string <code>s</code>, reconstruct the permutation <code>perm</code> and return it. If there are multiple valid permutations perm, return <strong>any of them</strong>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> s = "IDID"
<strong>Output:</strong> [0,4,1,3,2]
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> s = "III"
<strong>Output:</strong> [0,1,2,3]
</pre><p><strong class="example">Example 3:</strong></p>
<pre><strong>Input:</strong> s = "DDI"
<strong>Output:</strong> [3,2,0,1]
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 10<sup>5</sup></code></li>
	<li><code>s[i]</code> is either <code>&#39;I&#39;</code> or <code>&#39;D&#39;</code>.</li>
</ul>
Greedy; Array; Two Pointers; String
class Solution {
    public int[] diStringMatch(String s) {
        int n = s.length();
        int low = 0, high = n;
        int[] ans = new int[n + 1];
        for (int i = 0; i < n; i++) {
            if (s.charAt(i) == 'I') {
                ans[i] = low++;
            } else {
                ans[i] = high--;
            }
        }
        ans[n] = low;
        return ans;
    }
}