程式小試身手

上網查了不少資料,一直看到說使用getButton

但一直出錯顯示沒有getButton這方法,那時才意識到自己使用的不是AlertDialog 是 Dialog (自訂義 繼承AlertDialog.Builder)

將類型改為AlertDialog並在show()之後設定按鈕事件即可

import android.support.v7.app.AlertDialog;

AlertDialog.Builder builder = new AlertDialog.Builder(getActivity());
builder.setMessage("Test for preventing dialog close");
AlertDialog dialog = builder.create();
dialog.show();
//Overriding the handler immediately after show is probably a better approach than OnShowListener as described below
dialog.getButton(AlertDialog.BUTTON_POSITIVE).setOnClickListener(new View.OnClickListener()
      {            
          @Override
          public void onClick(View v)
          {
              Boolean wantToCloseDialog = false;
              //Do stuff, possibly set wantToCloseDialog to true then...
              if(wantToCloseDialog)
                  dismiss();
              //else dialog stays open. Make sure you have an obvious way to close the dialog especially if you set cancellable to false.
          }
      });

1. 生命週期

2. 内存泄漏？内存溢出 

3. string xml 用法

LinkedHashMap 概念非常不易 要另找時間再挑戰一次

有鑑於上一篇,有想到需要Hashmap,確實解答也是這麼用...

public List<String> wordBreak(String s, Set<String> wordDict) { return DFS(s, wordDict, new HashMap<String, LinkedList<String>>()); } // DFS function returns an array including all substrings derived from s. List<String> DFS(String s, Set<String> wordDict, HashMap<String, LinkedList<String>>map) { if (map.containsKey(s)) return map.get(s); LinkedList<String>res = new LinkedList<String>(); if (s.length() == 0) { res.add(""); return res; } for (String word : wordDict) { if (s.startsWith(word)) { List<String>sublist = DFS(s.substring(word.length()), wordDict, map); for (String sub : sublist) res.add(word + (sub.isEmpty() ? "" : " ") + sub); } } map.put(s, res); return res; }

能否從wordDict字典拚出字串s

想法一:

想到依序撈出字典去replace單字,並算長度是否為0,表示可以拼出

不過可能會有個誤區,可能先輪到的字典誤取代了字,所以我想在用同樣概念下循環字典就要不同順序了吧!

想法二:

Input: adjList = [[2,4],[1,3],[2,4],[1,3]]

Output: [[2,4],[1,3],[2,4],[1,3]]

一開始差點連題目都沒懂,原來是指1這個節點(也就是adjList的index 0)的連接點有2跟4

                                                    2這個節點(也就是adjList的index 1)的連接點有1跟3

我在mainfast.xml將Activity theme設定為NoActionBar

在Activity Java中定義了Toolbar 

參考Overstack作法: Toolbar Search Suggestions Theming

實作上一直以為自己出錯才沒有顯示提示,實際上是因為至少要輸入符合陣列內的內容!

特此筆記提醒自己

原本想先排序再Point index,但建議解真的很好,先將原index值+1 再去陣列循環是否包含此值,

以 [100,4,200,1,3,2] 來說 3+1=4,陣列確實含4且連續

class Solution {
     private boolean arrayContains(int[] arr, int num) {
        for (int i = 0; i < arr.length; i++) {
            System.out.println("arr=" + arr[i] );
            System.out.println("num=" + num );
            if (arr[i] == num) {
                return true;
            }
        }

        return false;
    }
    public int longestConsecutive(int[] nums) {
        int longestStreak = 0;

        for (int num : nums) {
            int currentNum = num;
            int currentStreak = 1;

class Solution {
    public int[][] insert(int[][] intervals, int[] newInterval) {
        List<int[]> ans = new ArrayList<>();
        int[] toAdd = newInterval;
        /* [[1,3],[6,9]]->[2,5] */
        for (int i = 0; i < intervals.length; i ++) {
            /*1. No overlap and toAdd appears before current interval, add toAdd to result.*/
            if (intervals[i][0] > toAdd[1]) {//1>5,6>5,1>3,6>3
                ans.add(toAdd);//[2,5],[1,3]
                toAdd = intervals[i];//[1,3],[6,9]
            }
            /*2. Has overlap, update the toAdd to the merged interval.*/
            else if (intervals[i][1] >= toAdd[0])  
                toAdd = new int[] {Math.min(intervals[i][0], toAdd[0]), Math.max(intervals[i][1], toAdd[1]) };
            /*3. No overlap and toAdd appears after current interval, add current interval to result.*/
            else ans.add(intervals[i]); 
        }
        ans.add(toAdd);
        return ans.toArray(new int[ans.size()][2]);
    }
}

挑了一個比較好懂得解答

覺得多數題目先判斷是否有值

// [[1,3],[2,6],[8,10],[15,18]] public int[][] merge(int[][] intervals) { int length=intervals.length;//4 if(length<=1) return intervals; int[] start = new int[length]; int[] end = new int[length]; for(int i=0;i<length;i++){ start[i]=intervals[i][0];//每個陣列第一個比[1,2,8,15] end[i]=intervals[i][1];//每個陣列第二個比[3,6,10,18] } Arrays.sort(start); Arrays.sort(end); int startIndex=0; int endIndex=0; List<int[]> result = new LinkedList<>(); while(endIndex<length){ //as endIndex==length-1 is evaluated first, start[endIndex+1] will never hit out of index if(endIndex==length-1 || start[endIndex+1]>end[endIndex]){//2>3->8>6->[1,6] result.add(new int[]{start[startIndex],end[endIndex]}); startIndex=endIndex+1; } endIndex++; } return result.toArray(new int[result.size()][]); }