Leetcode379 日常气温

Description

Given a list of daily temperatures T, return a list such 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 of temperatures T=[73,74,75,71,69,72,76,73] ,your output should be[1,1,4,2,1,1,0,0].

Note

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

Solution

一.我的解决办法
常规思路,遍历数组,如果后面有比今天气温还高的时候,返回下标的差值,否则,置0(自己写程序的时候很不仔细,这一部分没写对,导致试了好几遍才成功)。不置0的后果就是,应该为0的地方被后面不是0的位置的数字替代,而由于java数组定义的默认时候默认是0,后面本不该是0的地方都变成了0.

代码

class Solution {
    public int[] dailyTemperatures(int[] T) {
        int[] dailyTemperatures = new int [T.length];
        int m=0;
        for(int i=0;i<T.length;i++){
            int cur=T[i];
            for(int j=i+1;j<T.length;j++){
                if(T[j]>cur){
                  dailyTemperatures[m++]=j-i;                 
                break;}
                else{ 
                    if(j==T.length-1) dailyTemperatures[m++]=0;
                else{continue;}
                }
            }
            //dailyTemperatures[m]=0;
               
        }
        
    return dailyTemperatures;
        }
}

运行时间456ms

二.LeetCode Contributor的方法
思路:利用栈,栈中存的是给定的数组的下标,如果栈不为空并且栈顶元素小于数组中的元素,即弹出该下标;否则,入栈。

class Solution {
    public int[] dailyTemperatures(int[] temperatures) {
        final int m = temperatures.length;
        final int[] ans = new int[m];
        final Stack<Integer> stack = new Stack<>();
        for (int i = 0; i < m; i++) {
            while (!stack.empty() && temperatures[stack.peek()] < temperatures[i]) {
                ans[stack.peek()] = i - stack.pop();
            }
            stack.push(i);
        }
        return ans;
    }
}

运行时间56ms。
栈在本题中展示了强大的生命力,很有用。

附录

stack.peek()方法

猜你喜欢

转载自blog.csdn.net/weixin_42662955/article/details/89457956