LeetCode-739. Daily Temperatures

Description

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].

Solution 1(C++)

class Solution {
public:
    vector<int> dailyTemperatures(vector<int>& temperatures) {
        int len=temperatures.size();
        vector<int> res(len, 0);
        stack<int> s;
        for(int i=0; i<len; i++){
            while(!s.empty() && temperatures[i]>temperatures[s.top()]){
                int index = s.top();
                s.pop();
                res[index] = i - index;
            }
            s.push(i);
        }
        return res;
    }
};

Solution 2(C++)

class Solution {
public:
    vector<int> dailyTemperatures(vector<int>& temperatures) {
        vector<int> res(temperatures.size());
        for (int i = temperatures.size() - 1; i >= 0; --i) {
            int j = i+1;
            while (j < temperatures.size() && temperatures[j] <= temperatures[i]) {
                if (res[j] > 0) j = res[j] + j;
                else j = temperatures.size();
            }
            if (j < temperatures.size()) res[i] = j - i;
        }
        return res;
    }
};

Solution 3(C++)

class Solution {
public:
    vector<int> dailyTemperatures(vector<int>& temperatures) {
        int len=temperatures.size();
        vector<int> res(len, INT_MAX);
        unordered_map<int, int> m;
        for(int i=len-1; i>-1; i--){
            m[temperatures[i]] = i;
            int n=0;
            for(n=temperatures[i]+1; n<101; n++){
                if(m.find(n)!=m.end()) res[i]=min(res[i], m[n]-i);
            }
            if(res[i]==INT_MAX) res[i]=0;
        }
        return res;
    }
};

算法分析

其实题目就是要找到比当前temperatures[i]最近的大于它的数的坐标。

解法一： 如果一个子序列是递减的：i0，i1，i2，…，in，那么会一直往后查找，直到出现一个增加的数j，（这里的i与j都是索引）。然而这个变大的数也仅仅比之前的数in大。in对应的res[in]=1。如果j大于in-1，那么res[in-1]=2。依次类推。直到ik停止。那么ik~in的数对应res的值都确定了。剩下没有确定的自然就是i0~ik与新来的j。
注意： 这里j < ik。说明新加入的j还是满足递减的子序列的。所以没有确定的元素可以组成新的子序列，然后在temperatures继续往后遍历。

遇到这种需要储存“发生过”的信息，然后经过某些条件对储存信息进行更新，再加入新的信息，这种结构可以考虑stack与queue。由于这里“对储存信息进行更新”的那些信息优先选择是后进来的信息，所以符合stack的“先进先出”的特点，故而用stack比较好。

构建一个stack，用来储存递减元素的索引，当有新的元素要判断时，新元素与stack.top()比较，如果大于top，让res[stack.top()]=i-stack.top()，然后stack.pop()。重复以上过程，直到新元素小于stack.top()。将新元素压入栈。遍历完整个数组即可。

解法二： 参考：LeetCode-55. Jump Game 两者有一些地方相似。这个相似的地方比较抽象，我这里就不多说，可以多体会体会。这个解法是参考别人的答案：[C++] Clean code with explanation: O(n) time and O(1) space (beats 99.13%)。 别人答案写的很详细，我这里做一个备份：

We compute the result in reverse order. Given temperatures = [73, 74, 75, 71, 69, 72, 76, 73], suppose that we already obtain number of days for i >= 3: res = [?, ?, ?, 2, 1, 1, 0, 0] where ? denotes values to be determined. In order to find the number of days for i = 2, we proceed as follows:

1. Let j = i + 1, i.e., j = 3. If temperatures[i] < temperatures[j], res[i] = 1;
2. Otherwise, temperatures[i] >= temperatures[j]. we examine the value of res[j], which is equal to 2, which means that the most recent day with higher temperature than temperatures[j] in the future is 2 days away. Since temperatures[i] >= temperatures[j], the most recent day with higher temperature than temperatures[i] should be at least 2 days away. Therefore, we can skip some days and arrive directly at day j + res[j]. This process continues until we find higher temperature or we know we should stop as descibed in the third point.
3. If temperatures[j] == 0, we shoud directly set res[i] to 0 since we know we cannot find higher temperature in the future.

解法三： 考虑到温度的变化范围仅仅为：[30,100]，利用map储存temperatures中出现过温度的索引，当出现新的温度，可以在map中查找比该温度大的温度的索引。为了避免重复元素出现，可以从后往前遍历temperatures，并向map中添加新的索引。

程序分析

略。