diff --git a/Week01/Week01/NOTE.md b/Week01/Week01/NOTE.md
new file mode 100644
index 000000000..50de30414
--- /dev/null
+++ b/Week01/Week01/NOTE.md
@@ -0,0 +1 @@
+学习笔记
\ No newline at end of file
diff --git a/Week01/Week01/java/MoveZeroes.java b/Week01/Week01/java/MoveZeroes.java
new file mode 100644
index 000000000..ee6fc62f5
--- /dev/null
+++ b/Week01/Week01/java/MoveZeroes.java
@@ -0,0 +1,18 @@
+class Solution {
+    public void moveZeroes(int[] nums) {
+        int index = 0;
+        boolean b = false;
+        for(int i = 0; i <nums.length; i++){
+            if(nums[i]!=0&&b){
+                if(i!=index) {
+                    nums[index++] = nums[i];
+                    nums[i] = 0;
+                }
+            }else if(nums[i]==0){
+                b = true;
+            }else{
+                index++;
+            }
+        }
+    }
+}
\ No newline at end of file
diff --git a/Week01/Week01/java/RemoveDuplicates.java b/Week01/Week01/java/RemoveDuplicates.java
new file mode 100644
index 000000000..0085b8fb9
--- /dev/null
+++ b/Week01/Week01/java/RemoveDuplicates.java
@@ -0,0 +1,13 @@
+class Solution {
+    public int removeDuplicates(int[] nums) {
+        if(nums.length == 0)return 0;
+        int j = 0;
+        for(int i = 1; i < nums.length; i++){
+            if(nums[i]!=nums[j]){
+                j++;
+                nums[j] = nums[i];
+            }
+        }
+        return j+1;
+    }
+}
\ No newline at end of file
diff --git a/Week01/Week01/java/TwoNum.java b/Week01/Week01/java/TwoNum.java
new file mode 100644
index 000000000..b8bbc4781
--- /dev/null
+++ b/Week01/Week01/java/TwoNum.java
@@ -0,0 +1,12 @@
+class Solution {
+    public int[] twoSum(int[] nums, int target) {
+        Map<Integer,Integer> rMap = new HashMap<>(32);
+        for(int i =0;i<nums.length;i++){
+            if(rMap.containsKey(target-nums[i])){
+                return new int[]{rMap.get(target-nums[i]),i};
+            }
+            rMap.put(nums[i],i);
+        }
+        return new int[]{0,0};
+    }
+}
\ No newline at end of file
diff --git "a/Week01/Week01/\345\255\246\344\271\240\346\200\273\347\273\223.md" "b/Week01/Week01/\345\255\246\344\271\240\346\200\273\347\273\223.md"
new file mode 100644
index 000000000..8a4cbc3d1
--- /dev/null
+++ "b/Week01/Week01/\345\255\246\344\271\240\346\200\273\347\273\223.md"
@@ -0,0 +1,3 @@
+这周的内容是时间复杂度和线性数据结构：栈、队列、数组、链表
+数组和链表作为基础数据结构一般编程语言都通过这两种数据结构实现了栈、队列、跳表等数据结构
+学习知识还是需要更多的练习，对于链表尤其如此
\ No newline at end of file
diff --git a/Week04/NOTE.md b/Week04/NOTE.md
index 50de30414..e7d1100dc 100644
--- a/Week04/NOTE.md
+++ b/Week04/NOTE.md
@@ -1 +1,6 @@
-学习笔记
\ No newline at end of file
+week04学习笔记
+这周学习内容是深度优先搜索、广度优先搜索和贪心算法，还有二分查找
+内容很快,就到了第四周
+深度优先搜索基于递归思想，当然也可以通过自己手动实现一个栈来实现非递归的代码；广度和深度都需要一个visited数组来记录已经访问的节点；
+贪心算法代码实现一般比较简单，不过能想到贪心算法就需要不断地练习题目，来感受贪心的条件
+二分查找代码和链表的代码一样很容易写出错，也是需要大量的练习
diff --git a/Week04/java/LemonadeChange.java b/Week04/java/LemonadeChange.java
new file mode 100644
index 000000000..1237c49bd
--- /dev/null
+++ b/Week04/java/LemonadeChange.java
@@ -0,0 +1,35 @@
+public class LemonadeChange{
+    private boolean greedySolution(int[] bills) {
+        int five = 0, ten = 0;
+        for (int bill : bills) {
+            switch (bill) {
+                case 5: five++; break;
+                case 10: five--; ten++; break;
+                case 20: {
+                    if (ten > 0) {
+                        ten--; five--;
+                    } else {
+                        five -= 3;
+                    }
+                    break;
+                }
+                default: break;
+            }
+            if (five < 0) {
+                return false;
+            }
+        }
+        return true;
+    }
+
+}
+
+//在柠檬水摊上，每一杯柠檬水的售价为 5 美元。
+
+//顾客排队购买你的产品，（按账单 bills 支付的顺序）一次购买一杯。
+
+//每位顾客只买一杯柠檬水，然后向你付 5 美元、10 美元或 20 美元。你必须给每个顾客正确找零，也就是说净交易是每位顾客向你支付 5 美元。
+
+//注意，一开始你手头没有任何零钱。
+
+//如果你能给每位顾客正确找零，返回 true ，否则返回 false 。
diff --git a/Week04/java/NumberOfIslands.java b/Week04/java/NumberOfIslands.java
new file mode 100644
index 000000000..cea9ff9b9
--- /dev/null
+++ b/Week04/java/NumberOfIslands.java
@@ -0,0 +1,58 @@
+/**给你一个由 '1'（陆地）和 '0'（水）组成的的二维网格，请你计算网格中岛屿的数量。
+
+岛屿总是被水包围，并且每座岛屿只能由水平方向或竖直方向上相邻的陆地连接形成。
+
+此外，你可以假设该网格的四条边均被水包围。
+
+ 
+
+示例 1:
+
+输入:
+[
+['1','1','1','1','0'],
+['1','1','0','1','0'],
+['1','1','0','0','0'],
+['0','0','0','0','0']
+]
+输出: 1
+*/
+
+public class NumberOfIslands{
+     void dfs(char[][] grid, int r, int c) {
+            int nr = grid.length;
+            int nc = grid[0].length;
+
+            if (r < 0 || c < 0 || r >= nr || c >= nc || grid[r][c] == '0') {
+                return;
+            }
+
+            grid[r][c] = '0';
+            dfs(grid, r - 1, c);
+            dfs(grid, r + 1, c);
+            dfs(grid, r, c - 1);
+            dfs(grid, r, c + 1);
+        }
+
+        public int numIslands(char[][] grid) {
+            if (grid == null || grid.length == 0) {
+                return 0;
+            }
+
+            int nr = grid.length;
+            int nc = grid[0].length;
+            int num_islands = 0;
+            for (int r = 0; r < nr; ++r) {
+                for (int c = 0; c < nc; ++c) {
+                    if (grid[r][c] == '1') {
+                        ++num_islands;
+                        dfs(grid, r, c);
+                    }
+                }
+            }
+
+            return num_islands;
+        }
+}
+
+