739. Daily Temperatures

最新推荐文章于 2022-04-19 21:02:51 发布

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最新推荐文章于 2022-04-19 21:02:51 发布

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分类专栏： stack

本文链接：https://blog.youkuaiyun.com/u013527178/article/details/79297759

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本文介绍了一种高效的算法，用于预测每日气温何时会上升。通过使用栈或哈希表等数据结构，该算法能在O(n)的时间复杂度内完成计算，为用户提供未来温暖天气的等待天数。

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：

1.遍历每一个数，然后向后查找（超时）；

2.利用一个stack进行维护，获得一个新的数，比较其与stack.top()的大小，若大，pop出top的数，计算两个位置差存入，再与下一个比较，若小，直接压入stack

代码1：

class Solution {
public:
    vector<int> dailyTemperatures(vector<int>& temperatures) {
        vector< int >  ans ;
        if( temperatures.size() == 0 )
            return ans ;
        deque< int > dlist;
        deque< int > index ;
        ans.resize( temperatures.size() , 0 ) ;
        for( int i=0 ; i<temperatures.size() ; i++){
            if( dlist.size() == 0 ){
                dlist.push_back( temperatures[i] ) ;
                index.push_back( i );
            }
            else{
                if( temperatures[i] <= dlist.back() ){
                    dlist.push_back( temperatures[i] ) ;
                    index.push_back( i ) ;
                }
                else{
                    while( dlist.size() && dlist.back() < temperatures[i]) {
                        int target = index.back() ;
                        ans[target] = i - index.back() ;
                        index.pop_back() ;
                        dlist.pop_back() ;
                    }
                    dlist.push_back(temperatures[i]);
                    index.push_back( i ) ;
                }
            }
        }
        return ans  ;
    }
};

代码2：

class Solution {
public:
    vector<int> dailyTemperatures(vector<int>& t) {
        vector<int> res(t.size(),0);
        stack<pair<int,int>> s;
        for(int i=0;i<t.size();i++){
            pair<int,int> temp1(t[i],i);
            if(s.size()!=0){
                while(s.size()!=0){
                    pair<int,int> temp2=s.top();
                    if(temp2.first<temp1.first){
                        res[temp2.second]=temp1.second-temp2.second;
                        s.pop();
                    }
                    else{break;}
                }
            }
            s.push(temp1);
        }
        return res;
    }
};

代码3：

class Solution {
public:
    vector<int> dailyTemperatures(vector<int>& temps) {
        int n = temps.size();
        vector<int> waits(n, 0);
        vector<int> next(101, INT_MAX); // next day with with certain temperature.
        for (int i = n - 1; i >= 0; i--) {
            int earliest = INT_MAX;
            for (int t = temps[i] + 1; t <= 100; t++) {
                earliest = min(earliest, next[t]);
            }
            if (earliest != INT_MAX) waits[i] = earliest - i;
            next[temps[i]] = i;
        }
        return waits;
    }
};