<p>Given an array of integers <code>temperatures</code> represents the daily temperatures, return <em>an array</em> <code>answer</code> <em>such that</em> <code>answer[i]</code> <em>is the number of days you have to wait after the</em> <code>i<sup>th</sup></code> <em>day to get a warmer temperature</em>. If there is no future day for which this is possible, keep <code>answer[i] == 0</code> instead.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> temperatures = [73,74,75,71,69,72,76,73]
<strong>Output:</strong> [1,1,4,2,1,1,0,0]
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> temperatures = [30,40,50,60]
<strong>Output:</strong> [1,1,1,0]
</pre><p><strong class="example">Example 3:</strong></p>
<pre><strong>Input:</strong> temperatures = [30,60,90]
<strong>Output:</strong> [1,1,0]
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;=&nbsp;temperatures.length &lt;= 10<sup>5</sup></code></li>
	<li><code>30 &lt;=&nbsp;temperatures[i] &lt;= 100</code></li>
</ul>

class Solution {
public:
    vector<int> dailyTemperatures(vector<int>& temperatures) {
        stack<int> stk;
        int n = temperatures.size();
        vector<int> nextbig(n,0);
        for(int i=n-1;i>=0;i--){
            if(stk.empty()){
                stk.push(i);
                nextbig[i] = n;
            }
            else if(temperatures[i] < temperatures[stk.top()]){
                nextbig[i] = stk.top();
                stk.push(i);
            }
            else{
                while(!stk.empty() && temperatures[i] >= temperatures[stk.top()]){
                    stk.pop();
                }
                if(stk.empty()){
                    nextbig[i] = n;
                    stk.push(i);
                }
                else{
                    nextbig[i] = stk.top();
                    stk.push(i);
                }
            }
        }
        vector<int> ret(n,0);
        for(int i=0;i<n;i++){
            if(nextbig[i] == n){
                ret[i] = 0;
            }
            else{
                ret[i] = nextbig[i] - i;
            }
        }
        return ret;
    }
};
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