LeetCode刷题 | 739. Daily Temperatures

Given a list of daily temperatures, produce a list that, for each day in the input, tells you how many days you would have to wait until a warmer temperature. If there is no future day for which this is possible, put 0 instead.

For example, given the list temperatures = [73, 74, 75, 71, 69, 72, 76, 73], your output should be [1, 1, 4, 2, 1, 1, 0, 0].

Note: The length of temperatures will be in the range [1, 30000]. Each temperature will be an integer in the range [30, 100].

思路：这题要求给出一个数组，求出数组中每一个数其之后第一个比它大的数字离它有多远，这题可以暴力解决，不过这样时间肯定很长，可以使用一种类似线性DP的方法去解决，

假设该数组为num[0],num[1],…num[n],我们用一个ob[]数组，ob[k]表示比num[k]大的第一个数字在num[]中的位置，假设右为正方向，我们从右向左计算ob[]的值，也就是说，当计算ob[k]的时候，ob[k]-ob[n]都已经得到了正确的值，假设num[k]比num[k+1]要小，毫无疑问ob[k]=k+1;假如num[k]>=num[k+1],那么比较num[k]和num[ob[k+1]]的大小，如果num[k]>= num[ob[k+1]],那么比较num[k]和num[ob[ob[k+1]]]的大小。。。就这样一直比较，假如某一次发现num[ob[ob[…k+1]]…]为-1（-1表示没有比它大的数），那么ob[k]为-1

class Solution {

public:

    vector<int>dailyTemperatures(vector<int>& temperatures) {

       vector <int> a; vector <int>b;

      for (int i = temperatures.size() - 1; i>= 0; i--)

      {

            if (i + 1<temperatures.size())

            {

                  if(temperatures[i]<temperatures[i + 1]) a.push_back(1);

                  else

                  {

                       int index =temperatures.size() - 1 - i - 1; int count = 1;

                       int num = a[index]; inttemIndex = i + 1;

                       while (num != 0)

                       {

                             temIndex = temIndex+ num;

                             if(temperatures[temIndex]>temperatures[i]) { count += num; a.push_back(count);break; }

                             else { count +=num; index = temperatures.size() - 1 - temIndex; num = a[index]; }

                       }

                       if(num == 0)

                       a.push_back(0);

                  }

            }

            else

            {

                  a.push_back(0);

            }

      }

      for (int i = a.size() - 1; i >= 0; i--)

      {

            b.push_back(a[i]);

      }

         return b;

    }

};