# Leetcode 739. Daily Temperatures
###### tags: `Leetcode`

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

Example 1:

Input: temperatures = [73,74,75,71,69,72,76,73]
Output: [1,1,4,2,1,1,0,0]
Example 2:

Input: temperatures = [30,40,50,60]
Output: [1,1,1,0]
Example 3:

Input: temperatures = [30,60,90]
Output: [1,1,0]

解法:
1.把當下的溫度值拿來和存在stack的值比較，若有比較大，就把那個pop出來，然後當下的值就直接push進去，等待後面的審查
2.原本擔心會有先push進去的就不會被檢查到的問題，但根本不會發生，因為如果最上層的都出不去，那表示他下面的也一定都比他小，所以就檢查stack的最上層就好。喔對，難怪是要用stack，這題就是後進先出的概念= =。
3.malloc沒初始化、calloc可以初始化

/**
 

/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
int* dailyTemperatures(int* temperatures, int temperaturesSize, int* returnSize){
    *returnSize = temperaturesSize;
    int* res = calloc(temperaturesSize, sizeof(int));
    int stack[temperaturesSize];
    int stackSize = 0;//index of stack = stackSize - 1
    for(int i = 0; i < temperaturesSize; i++)
    {
        while(stackSize != 0 && temperatures[i] > temperatures[stack[stackSize-1]])
        {
            res[stack[stackSize-1]] = i - stack[stackSize-1];
            stackSize--;
        }
        stack[stackSize] = i;
        stackSize++;
    }
    return res;
}