IncredibleThings

导航

LeetCode - 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].

竟然暴力解法也可AC:

class Solution {
    public int[] dailyTemperatures(int[] temperatures) {
        if(temperatures == null || temperatures.length == 0){
            return null;
        }
        int len = temperatures.length;
        int[] res = new int[len];
        for(int i=0; i<len; i++){
            int count = 0;
            int cur = temperatures[i];
            for(int j = i+1; j<len; j++){
                int fu = temperatures[j];
                if(fu > cur){
                    count = j-i;
                    break;
                }
            }
            res[i] = count;
        }
        return res;
    }
}

  另一种使用stack的解法:

class Solution {
    public int[] dailyTemperatures(int[] temperatures) {
        if(temperatures == null || temperatures.length == 0){
            return null;
        }
        int len = temperatures.length;
        int[] res = new int[len];
        Stack<Integer> stack = new Stack<>();
        for(int i = 0; i<len; i++){
            while(!stack.isEmpty() && temperatures[i] > temperatures[stack.peek()]){
                int temp = stack.pop();
                res[temp] = i - temp;
            }
            stack.push(i);
            
        }
        return res;
    }
}

 

posted on 2018-09-23 08:21  IncredibleThings  阅读(93)  评论(0编辑  收藏  举报