m crush

Author: awangdev

Java/Merge Sorted Array II.java

@@ -1,7 +1,15 @@
-长度已经固定。普通做法。
+E
+1528179898
+tags: Array
+
+如题, merge two sorted array into 新的 sorted array
+
+- 长度已经固定. Basic Implementation
+- 如果一个array足够大, merge into this array, 那么就是从末尾merge.
+
 ```
 /*
-33% Accepted
+
 Merge two given sorted integer array A and B into a new sorted integer array.
 
 Example
@@ -26,10 +34,6 @@
 */
 
 class Solution {
-    /**
-     * @param A and B: sorted integer array A and B.
-     * @return: A new sorted integer array
-     */
     public int[] mergeSortedArray(int[] A, int[] B) {
         if (A == null || B == null) {
             return A == null ? B : A;

Java/Minimum Subarray.java

@@ -1,3 +1,10 @@
+E
+tags: Greedy, Array
+
+TODO: dp?
+
+```
+
 /*
 Given an array of integers, find the subarray with smallest sum.
 
@@ -40,4 +47,5 @@ public int minSubArray(ArrayList<Integer> nums) {
 		}
 		return minRst;
     }
-}
+}
+```

Java/Nth to Last Node in List.java

@@ -1,7 +1,12 @@
-先找到nth node
-然后head开始跑。
+E
+1528180250
+tags: Linked List
+
+#### Linked List
+- 先找到nth node
+- 然后head开始跑
+- node 到底，而head ~ node刚好是 n 距离。所以head就是要找的last nth
 
-node 到底，而head ~ node刚好是 n 距离。所以head就是要找的last nth
 ```
 /*
 Find the nth to last element of a singly linked list. 

Java/O(1) Check Power of 2.java

@@ -1,3 +1,8 @@
+E
+tags: Bit Manipulation
+
+```
+
 /*
 Using O(1) time to check whether an integer n is a power of 2.
 Example
@@ -32,3 +37,5 @@ public boolean checkPowerOf2(int n) {
 };
 
 
+
+```

Java/Partition Array by Odd and Even.java

@@ -1,3 +1,10 @@
+E
+tags: Two Pointers, Array
+
+- 更正常的start/end partition pointer类似: when condition meet, swap
+- Clean up TODO
+
+```
 /*
 Partition an integers array into odd number first and even number second.
 
@@ -49,3 +56,5 @@ public void partitionArray(int[] nums) {
     	}
     }
 }
+
+```

Java/Product of Array Exclude Itself.java

@@ -1,4 +1,31 @@
+M
+tags: Array
+
+
+```
+/*
+LeetCode:
+Given an array nums of n integers where n > 1,  
+return an array output such that output[i] is equal to the product of all the elements of nums except nums[i].
+
+Example:
+
+Input:  [1,2,3,4]
+Output: [24,12,8,6]
+Note: Please solve it without division and in O(n).
+
+Follow up:
+Could you solve it with constant space complexity? 
+(The output array does not count as extra space for the purpose of space complexity analysis.)
+
+*/
+
+
+
+
+
 /*
+LintCode
 Given an integers array A.
 
 Define B[i] = A[0] * ... * A[i-1] * A[i+1] * ... * A[n-1], calculate B WITHOUT divide operation.
@@ -42,3 +69,5 @@ public ArrayList<Long> productExcludeItself(ArrayList<Integer> A) {
 }
 
 
+
+```

Java/Recover Rotated Sorted Array.java

@@ -1,3 +1,6 @@
+E
+tags: Array
+
 rotate的意思，是有个点断开，把一边的array节选出来放在另外一边。
 Rotate三步：
 rotate前半

Java/Remove Duplicates from Unsorted List.java

@@ -1,3 +1,6 @@
+M
+tags: Linked List
+
 基本方法: O(n) sapce, time
 遍历。
 遇到duplicate(可能多个),  while直到node.next不是duplicate.

Java/Search Range in Binary Search Tree .java

@@ -1,11 +1,13 @@
 M
-tags: BST
+1528177257
+tags: BST, Binary Tree
 
-等于遍历了所有k1<= x <= k2的x node。
+给一个BST, integer range (k1, k2), 找range 里面所有的integer.
 
-如果是用Binary Search Tree搜索，那么一般是if (...) else {...}，也就是一条路走到底，直到找到target.
-
-这里, 把 left/right/match的情况全部cover了，然后把k1,k2的边框限制好，中间就全部遍历了。
+#### BST
+- 等于dfs遍历了所有k1<= x <= k2的x node。
+- dfs left, process root, then dfs right
+- 这里, 把 left/right/match的情况全部cover了，然后把k1,k2的边框限制好，中间就全部遍历了。
 
 ```
 /*
@@ -44,17 +46,18 @@ Given two values k1 and k2 (where k1 < k2) and a root pointer to a Binary Search
 
 public class Solution {
     /**
-     * @param root: The root of the binary search tree.
-     * @param k1 and k2: range k1 to k2.
-     * @return: Return all keys that k1<=key<=k2 in ascending order.
+     * @param root: param root: The root of the binary search tree
+     * @param k1: An integer
+     * @param k2: An integer
+     * @return: return: Return all keys that k1<=key<=k2 in ascending order
      */
-    public ArrayList<Integer> searchRange(TreeNode root, int k1, int k2) {
-        ArrayList<Integer> result = new ArrayList<Integer>();
+    public List<Integer> searchRange(TreeNode root, int k1, int k2) {
+        List<Integer> result = new ArrayList<>();
         helper(result, root, k1, k2);
         return result;
     }
 
-    public void helper(ArrayList<Integer> result, TreeNode root, int k1, int k2) {
+    public void helper(List<Integer> result, TreeNode root, int k1, int k2) {
         if (root == null) {
             return;
         }
@@ -70,5 +73,4 @@ public void helper(ArrayList<Integer> result, TreeNode root, int k1, int k2) {
     }
 }
 
-
 ```

Java/Search Rotated in Sorted Array II.java

@@ -1,9 +1,39 @@
+M
+tags: Array, Binary Search
+
 Allow duplicates之后：
 因为最终binary search的结果也是O(n)
 所以这道题要记得： 既然是O(n), 那来个简单的for loop 也就好了。
 
 当然，要跟面试官提起来原因。别一上来就只有for。。。
 ```
+/*
+LeetCode
+Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.
+
+(i.e., [0,0,1,2,2,5,6] might become [2,5,6,0,0,1,2]).
+
+You are given a target value to search. If found in the array return true, otherwise return false.
+
+Example 1:
+
+Input: nums = [2,5,6,0,0,1,2], target = 0
+Output: true
+Example 2:
+
+Input: nums = [2,5,6,0,0,1,2], target = 3
+Output: false
+Follow up:
+
+This is a follow up problem to Search in Rotated Sorted Array, where nums may contain duplicates.
+Would this affect the run-time complexity? How and why?
+
+
+*/
+
+
+
+
 /*
 Follow up for "Search in Rotated Sorted Array":
 What if duplicates are allowed?

Java/Search a 2D Matrix II.java

@@ -1,9 +1,95 @@
 M
+1528169466
+tags: Binary Search, Divide and Conquer
+
+给matrix, 每一行sorted, 每一列从上往下sorted, 找target是否存在
+
+#### Binary Search
+- 根据给定的性质, 其实点选的极端一点: x = 最下面的row, y = 当下一行里面最小的left position. 
+- (x,y)在左下角
+- 在此情况下, 只能往一个方向运行: 如果小于target, y++; 如果大于target, 那么只能x--
+- 每次操作, 都是删掉一行, 或者一列, 再也不需要回头看
+- `while (x >= 0 && y < col) {}` 确保不会跑脱
+- 同样的方式: 可以从右上角(0, col - 1) 开始, 代码稍微改一改
+
+#### Divide and Conquer?
+- TODO
 
-根据给定的性质，找个点（left-bottom）：每次有任何情况，只能往一个方向运行。
-每次删掉一行，或者一列
 ```
+/**
+LeetCode: check existance
+Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:
+
+Integers in each row are sorted in ascending from left to right.
+Integers in each column are sorted in ascending from top to bottom.
+Consider the following matrix:
+
+[
+  [1,   4,  7, 11, 15],
+  [2,   5,  8, 12, 19],
+  [3,   6,  9, 16, 22],
+  [10, 13, 14, 17, 24],
+  [18, 21, 23, 26, 30]
+]
+Example 1:
+
+Input: matrix, target = 5
+Output: true
+Example 2:
+
+Input: matrix, target = 20
+Output: false
+*/
+// from bottom-left, (x,y) = (row -1, 0)
+public class Solution {
+    public boolean searchMatrix(int[][] matrix, int target) {
+        if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
+            return false;
+        }
+        int row = matrix.length;
+        int col = matrix[0].length;
+        int x = row - 1;
+        int y = 0;
+        
+        while (x >= 0 && y < col) {
+            if (matrix[x][y] < target) {
+                y++;
+            } else if (matrix[x][y] > target) {
+                x--;
+            } else {//matrix[x][y] == target
+                return true;
+            }
+        }
+        return false;
+    }
+}
+
+// from top-right, (x,y) = (0, col - 1)
+public class Solution {
+    public boolean searchMatrix(int[][] matrix, int target) {
+        if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
+            return false;
+        }
+        int row = matrix.length;
+        int col = matrix[0].length;
+        int x = 0;
+        int y = col - 1;
+        
+        while (x < row && y >= 0) {
+            if (matrix[x][y] < target) {
+                x++;
+            } else if (matrix[x][y] > target) {
+                y--;
+            } else {//matrix[x][y] == target
+                return true;
+            }
+        }
+        return false;
+    }
+}
+
 /*
+LintCode: find # of occurances
 Write an efficient algorithm that searches for a value in an m x n matrix, return the occurrence of it.
 
 This matrix has the following properties:

Java/Search a 2D Matrix.java

@@ -1,8 +1,13 @@
 M
+1528167745
+tags: Array, Binary Search
+
+给2D matrix, 每行sorted, 每行的首位都大于上一行的末尾. goal: find target from matrix
+
+#### 2D matrix 转1D array
+- 一行一行是从小到大, sorted, 连续的, 可以看做1D sorted array
+- Binary Search
 
-一行一行是从小到大，连续的：
-2D转1D。
-Binary Search
 ```
 /*
 Write an efficient algorithm that searches for a value in an m x n matrix.