好啦，今天國慶日，我不想逼迫自己咩，其實是我今天寫不出來，寫不出的題號是56. Merge Intervals，我一直覺得差一點點了，然後家裡沒有書桌可以用真的是很崩潰ㄟ，我晚上如果有時間再試試看寫不寫得出來吧，等等吃燒烤去
好啦，今天國慶日，我不想逼迫自己咩，其實是我今天寫不出來，寫不出的題號是56. Merge Intervals，我一直覺得差一點點了，然後家裡沒有書桌可以用真的是很崩潰ㄟ，我晚上如果有時間再試試看寫不寫得出來吧，等等吃燒烤去

目前的程式碼

class Solution {
    public int[][] merge(int[][] intervals) {
        int[] temp = new int[2];
        Arrays.fill(temp, 0);
        List<List<Integer>> rtemp = new ArrayList<List<Integer>>();
        List<List<Integer>> rtemp2 = new ArrayList<List<Integer>>();
        int i,j,k=0,a=-1,b=-1;
        
        for ( i = 0; i < intervals.length; i++) {
            Arrays.sort(intervals[i]);
        }
        for ( i = 0; i < intervals.length; i++) {
            System.out.println(intervals[i][0] +" " +intervals[i][1]);
        }
        
        for(i=0;i<intervals.length;i++){
           if(rtemp2.contains(Arrays.asList(intervals[i][0],intervals[i][1]))){
           }else{
               rtemp2.add(Arrays.asList(intervals[i][0],intervals[i][1]));
           }
        }
       
        temp[0] = rtemp2.get(0).get(0);
        temp[1] = rtemp2.get(0).get(1);
        for(i=1;i<rtemp2.size();i++){
            a = rtemp2.get(i).get(0);
            b = rtemp2.get(i).get(1);
            if(a>temp[1]){
                rtemp.add(Arrays.asList(temp[0],temp[1]));
                temp[0] = rtemp2.get(i).get(0);
                temp[1] = rtemp2.get(i).get(1);
            }else if(a<=temp[1]&&b>=temp[1]){
                temp[1]= b;
            }
        }
 
        if(!rtemp.contains(Arrays.asList(temp[0],temp[1]))){
            rtemp.add(Arrays.asList(temp[0],temp[1]));
        }
        
        //rtemp.add(Arrays.asList(temp[0],temp[1]));
        int[][] result = new int[rtemp.size()][2];
        for(i=0;i<rtemp.size();i++){
           for(j=0;j<2;j++){
               result[i][0] = rtemp.get(i).get(0);
               result[i][1] = rtemp.get(i).get(1);
           }
        }
        return result;
    }
}

sort的部分還沒處理好，之後再處理囉

二更)

我竟然解決啦

題目:56 標題:Merge Intervals 難度:Medium

Given an array of intervals where intervals[i] = [starti, endi], merge all overlapping intervals, and return an array of the non-overlapping intervals that cover all the intervals in the input.

Example 1:
Input: intervals = [[1,3],[2,6],[8,10],[15,18]]
Output: [[1,6],[8,10],[15,18]]
Explanation: Since intervals [1,3] and [2,6] overlaps, merge them into [1,6].

Example 2:
Input: intervals = [[1,4],[4,5]]
Output: [[1,5]]
Explanation: Intervals [1,4] and [4,5] are considered overlapping.

Constraints:
1 <= intervals.length <= 104
intervals[i].length == 2
0 <= starti <= endi <= 104

成功的程式碼

class Solution {
    public int[][] merge(int[][] intervals) {
        int[] temp = new int[2];
        Arrays.fill(temp, 0);
        List<List<Integer>> rtemp = new ArrayList<List<Integer>>();
        List<List<Integer>> rtemp2 = new ArrayList<List<Integer>>();
        int i,j,k=0,a=-1,b=-1;
        
        for ( i = 0; i < intervals.length; i++) {
            Arrays.sort(intervals, (x, y) -> x[0] - y[0]);
        }
        for ( i = 0; i < intervals.length; i++) {
            System.out.println(intervals[i][0] +" " +intervals[i][1]);
        }
        
        for(i=0;i<intervals.length;i++){
           if(rtemp2.contains(Arrays.asList(intervals[i][0],intervals[i][1]))){
           }else{
               rtemp2.add(Arrays.asList(intervals[i][0],intervals[i][1]));
           }
        }
       
        temp[0] = rtemp2.get(0).get(0);
        temp[1] = rtemp2.get(0).get(1);
        for(i=1;i<rtemp2.size();i++){
            a = rtemp2.get(i).get(0);
            b = rtemp2.get(i).get(1);
            if(a>temp[1]){
                rtemp.add(Arrays.asList(temp[0],temp[1]));
                temp[0] = rtemp2.get(i).get(0);
                temp[1] = rtemp2.get(i).get(1);
            }else if(a<=temp[1]&&b>=temp[1]){
                temp[1]= b;
            }
        }
 
        if(!rtemp.contains(Arrays.asList(temp[0],temp[1]))){
            rtemp.add(Arrays.asList(temp[0],temp[1]));
        }
        
        int[][] result = new int[rtemp.size()][2];
        for(i=0;i<rtemp.size();i++){
           for(j=0;j<2;j++){
               result[i][0] = rtemp.get(i).get(0);
               result[i][1] = rtemp.get(i).get(1);
           }
        }
        return result;
    }
}

DAY25心得
明天得再寫一題了QQ