finish 178, 566

BaffinLee · BaffinLee · commit cc72d82c5882 · 2018-08-14T18:59:53.000+08:00
diff --git a/101-200/177. Nth Highest Salary.md b/101-200/177. Nth Highest Salary.md
@@ -35,7 +35,7 @@ CREATE FUNCTION getNthHighestSalary(N INT) RETURNS INT
 BEGIN
   RETURN (
       # Write your MySQL query statement below.
-      SELECT DISTINCT Salary FROM Employee ORDER BY Salary DESC LIMIT M, 1
+      SELECT DISTINCT Salary FROM Employee ORDER BY Salary DESC LIMIT N, 1
   );
 END
 ```
diff --git a/101-200/178. Rank Scores.md b/101-200/178. Rank Scores.md
@@ -0,0 +1,81 @@
+# 178. Rank Scores
+
+- Difficulty: Medium.
+- Related Topics: .
+- Similar Questions: .
+
+## Problem
+
+Write a SQL query to rank scores. If there is a tie between two scores, both should have the same ranking. Note that after a tie, the next ranking number should be the next consecutive integer value. In other words, there should be no "holes" between ranks.
+
+```
++----+-------+
+| Id | Score |
++----+-------+
+| 1  | 3.50  |
+| 2  | 3.65  |
+| 3  | 4.00  |
+| 4  | 3.85  |
+| 5  | 4.00  |
+| 6  | 3.65  |
++----+-------+
+```
+
+For example, given the above ```Scores``` table, your query should generate the following report (order by highest score):
+
+```
++-------+------+
+| Score | Rank |
++-------+------+
+| 4.00  | 1    |
+| 4.00  | 1    |
+| 3.85  | 2    |
+| 3.65  | 3    |
+| 3.65  | 3    |
+| 3.50  | 4    |
++-------+------+
+```
+
+## Solution 1
+
+```javascript
+# Write your MySQL query statement below
+SELECT
+  Score,
+  (SELECT count(distinct Score) FROM Scores WHERE Score >= s.Score) as Rank
+FROM
+  Scores s
+ORDER BY
+  Score DESC
+```
+
+**Explain:**
+
+nope.
+
+**Complexity:**
+
+* Time complexity :
+* Space complexity :
+
+## Solution 2
+
+```javascript
+# Write your MySQL query statement below
+SELECT
+  Score,
+  (SELECT count(*) FROM (SELECT distinct Score as s FROM Scores) tmp WHERE s >= Score) as Rank
+FROM
+  Scores
+ORDER BY
+  Score DESC
+```
+
+**Explain:**
+
+nope.
+
+**Complexity:**
+
+* Time complexity :
+* Space complexity :
diff --git a/501-600/566. Reshape the Matrix.md b/501-600/566. Reshape the Matrix.md
@@ -0,0 +1,80 @@
+# 566. Reshape the Matrix
+
+- Difficulty: Easy.
+- Related Topics: Array.
+- Similar Questions: .
+
+## Problem
+
+In MATLAB, there is a very useful function called 'reshape', which can reshape a matrix into a new one with different size but keep its original data.
+
+You're given a matrix represented by a two-dimensional array, and two **positive** integers **r** and **c** representing the **row** number and **column** number of the wanted reshaped matrix, respectively.
+
+The reshaped matrix need to be filled with all the elements of the original matrix in the same **row-traversing** order as they were.
+
+If the 'reshape' operation with given parameters is possible and legal, output the new reshaped matrix; Otherwise, output the original matrix.
+
+**Example 1:**
+```
+Input: 
+nums = 
+[[1,2],
+ [3,4]]
+r = 1, c = 4
+Output: 
+[[1,2,3,4]]
+Explanation:The row-traversing of nums is [1,2,3,4]. The new reshaped matrix is a 1 * 4 matrix, fill it row by row by using the previous list.
+```
+
+**Example 2:**
+```
+Input: 
+nums = 
+[[1,2],
+ [3,4]]
+r = 2, c = 4
+Output: 
+[[1,2],
+ [3,4]]
+Explanation:There is no way to reshape a 2 * 2 matrix to a 2 * 4 matrix. So output the original matrix.
+```
+
+**Note:**
+
+- The height and width of the given matrix is in range [1, 100].
+- The given r and c are all positive.
+
+## Solution
+
+```javascript
+/**
+ * @param {number[][]} nums
+ * @param {number} r
+ * @param {number} c
+ * @return {number[][]}
+ */
+var matrixReshape = function(nums, r, c) {
+  var m = nums.length;
+  var n = nums[0].length;
+  
+  if (m * n !== r * c) return nums;
+  
+  var res = Array(r).fill(0).map(_ => Array(c));
+  var num = r * c;
+  
+  for (var i = 0; i < num; i++) {
+    res[Math.floor(i / c)][i % c] = nums[Math.floor(i / n)][i % n];
+  }
+
+  return res;
+};
+```
+
+**Explain:**
+
+nope.
+
+**Complexity:**
+
+* Time complexity : O(r*c).
+* Space complexity : O(r*c).
