0034、0035、0036 solved

0034-find-first-and-last-position-of-element-in-sorted-array/Article/0034-find-first-and-last-position-of-element-in-sorted-array.md

@@ -0,0 +1,81 @@
+# LeetCode 第 34 号问题：在排序数组中查找元素的第一个和最后一个位置
+
+
+题目来源于 LeetCode 上第 34 号问题：find-first-and-last-position-of-element-in-sorted-array。题目难度为 中等。
+
+### 题目描述
+
+给定一个按照升序排列的整数数组 **nums**，和一个目标值 **target**。找出给定目标值在数组中的开始位置和结束位置。
+
+你的算法时间复杂度必须是 **O(log n)** 级别。
+
+如果数组中不存在目标值，返回 [-1, -1]。
+
+
+**示例:**
+
+```
+输入: nums = [5,7,7,8,8,10], target = 8
+输出: [3,4]
+```
+```
+输入: nums = [5,7,7,8,8,10], target = 6
+输出: [-1,-1]
+```
+### 题目解析
+
+题目中要求了时间复杂度为O(log n),这就很清楚要使用二分查找法了。
+
+首先定义两个指针变量，分别存储左右两个位置的索引。首先去找目标值的最左面的索引,通过循环为了防止元素丢失,每次保留最右面的元素,左侧的指针移动时+1。在循环结束的时候判断一下数组中是否包括目标值,不包括的话直接退出。
+右面的跟左侧相同，只不过正好相反。
+
+
+
+### 动画描述
+
+![](..\Animation\在排序数组中查找元素的第一个和最后一个位置.gif)
+
+### 代码实现
+
+```java
+// 34. 下一个排列
+// https://leetcode-cn.com/problems/find-first-and-last-position-of-element-in-sorted-array/
+// 时间复杂度：O(n)
+// 空间复杂度：O(1)
+class Solution {
+    public int[] searchRange(int[] nums, int target) {
+		int[] res = new int[] { -1, -1 };
+		int left = 0;
+		int right = nums.length - 1;
+		int l = left;
+		int r = right;
+		while (left < right) {
+			int mid = (left + right) / 2;
+			if (nums[mid] < target) {
+				left = mid + 1;
+			} else {
+				right = mid;
+			}
+		}
+		if (left>right||nums[left]!=target) {
+			return new int[]{-1,-1};
+		}
+		while (l < r) {
+			int mid = (l + r) / 2 + 1;
+			if (nums[mid] > target) {
+				r = mid - 1;
+			} else {
+				l = mid;
+			}
+		}
+		if (left > right || left > r) {
+			return new int[] { -1, -1 };
+		} else {
+			return new int[] { left, r };
+		}
+	}
+}
+
+```
+
+ ![](../../Pictures/qrcode.jpg)

0034-find-first-and-last-position-of-element-in-sorted-array/Code/1.java

@@ -0,0 +1,33 @@
+class Solution {
+    public int[] searchRange(int[] nums, int target) {
+		int[] res = new int[] { -1, -1 };
+		int left = 0;
+		int right = nums.length - 1;
+		int l = left;
+		int r = right;
+		while (left < right) {
+			int mid = (left + right) / 2;
+			if (nums[mid] < target) {
+				left = mid + 1;
+			} else {
+				right = mid;
+			}
+		}
+		if (left>right||nums[left]!=target) {
+			return new int[]{-1,-1};
+		}
+		while (l < r) {
+			int mid = (l + r) / 2 + 1;
+			if (nums[mid] > target) {
+				r = mid - 1;
+			} else {
+				l = mid;
+			}
+		}
+		if (left > right || left > r) {
+			return new int[] { -1, -1 };
+		} else {
+			return new int[] { left, r };
+		}
+	}
+}

0035-search-insert-position/Article/0035-search-insert-position.md

@@ -0,0 +1,113 @@
+# LeetCode 第 35 号问题：搜索插入位置
+
+> 本文首发于公众号「图解面试算法」，是 [图解 LeetCode ](<https://github.com/MisterBooo/LeetCodeAnimation>) 系列文章之一。
+>
+> 同步博客：https://www.algomooc.com
+
+题目来源于 LeetCode 第 35 号问题：搜索插入位置. 
+
+## 题目
+
+给定一个排序数组和一个目标值，在数组中找到目标值，并返回其索引。如果目标值不存在于数组中，返回它将会被按顺序插入的位置。
+你可以假设数组中无重复元素。
+
+
+示例 1:
+
+```
+输入: [1,3,5,6], 5
+输出: 2
+```
+
+示例 2:
+
+
+```
+输入: [1,3,5,6], 2
+输出: 1
+```
+
+示例 3:
+
+
+```
+输入: [1,3,5,6], 7
+输出: 4
+```
+
+
+示例 4:
+
+
+```
+输入: [1,3,5,6], 0
+输出: 0
+```
+
+
+
+## 思路解析
+
+### 暴力循环法
+
+这个题看起来就是很简单的，就是一道考验查找算法的题目。最简单的就是暴力查找了。
+
+#### 思路
+
+遍历这个数组，然后如果当前值和目标值target一致或小于目标值target，那么就return 当前下标。这种解法的时间复杂度是O(N)
+
+###  动画理解
+
+![](../Animation/暴力查找.gif)
+
+#### 代码实现
+
+
+```java
+//时间复杂度：O(n)
+//空间复杂度：O(1)
+class Solution {
+    public int searchInsert(int[] nums, int target) {
+        int i=0;
+        for(;i<nums.length;i++){
+            if (nums[i]>=target){
+                break;
+            }
+        }
+        return i;
+    }
+}
+```
+
+### 二分法
+
+#### 思路
+
+除了暴力法，我们在排序数组中查找值还可以用的一种方法是二分法，思路还是和改良的暴力循环法一样，先找到左右边界，然后计算，每次可以省出一半的时间。时间复杂度为O(logn)
+
+#### 代码实现
+
+```java
+//时间复杂度：O(lon(n))
+//空间复杂度：O(1)
+class Solution {
+    public int searchInsert(int[] nums, int target) {
+        if (target>nums[nums.length-1]) {
+            return nums.length;
+        }
+        int left=0;
+        int right=nums.length-1;
+        while (left < right) {
+            int mid = (left + right) / 2;
+            if (nums[mid] < target) {
+                left = mid + 1;
+            } else {
+                right = mid;
+            }
+        }
+        return left;
+
+    }
+}
+```
+ ![](../../Pictures/qrcode.jpg)

0035-search-insert-position/Code/1.java

@@ -0,0 +1,11 @@
+class Solution1 {
+    public int searchInsert(int[] nums, int target) {
+        int i=0;
+        for(;i<nums.length;i++){
+            if (nums[i]>=target){
+                break;
+            }
+        }
+        return i;
+    }
+}

0035-search-insert-position/Code/2.java

@@ -0,0 +1,21 @@
+//时间复杂度：O(lon(n))
+//空间复杂度：O(1)
+class Solution2 {
+    public int searchInsert(int[] nums, int target) {
+        if (target>nums[nums.length-1]) {
+            return nums.length;
+        }
+        int left=0;
+        int right=nums.length-1;
+        while (left < right) {
+            int mid = (left + right) / 2;
+            if (nums[mid] < target) {
+                left = mid + 1;
+            } else {
+                right = mid;
+            }
+        }
+        return left;
+
+    }
+}

0036-valid-sudoku/Article/0036-valid-sudoku.md

@@ -0,0 +1,134 @@
+# LeetCode 第 36 号问题：有效的数独
+
+> 本文首发于公众号「图解面试算法」，是 [图解 LeetCode ](<https://github.com/MisterBooo/LeetCodeAnimation>) 系列文章之一。
+>
+> 同步博客：https://www.algomooc.com
+
+题目来源于 LeetCode 第 36 号问题：有效的数独. 
+
+## 题目
+
+判断一个 9x9 的数独是否有效。只需要根据以下规则，验证已经填入的数字是否有效即可。
+
+
+	数字 1-9 在每一行只能出现一次。
+	数字 1-9 在每一列只能出现一次。
+	数字 1-9 在每一个以粗实线分隔的 3x3 宫内只能出现一次。
+
+
+示例 1:
+
+```
+输入:
+[
+  ["5","3",".",".","7",".",".",".","."],
+  ["6",".",".","1","9","5",".",".","."],
+  [".","9","8",".",".",".",".","6","."],
+  ["8",".",".",".","6",".",".",".","3"],
+  ["4",".",".","8",".","3",".",".","1"],
+  ["7",".",".",".","2",".",".",".","6"],
+  [".","6",".",".",".",".","2","8","."],
+  [".",".",".","4","1","9",".",".","5"],
+  [".",".",".",".","8",".",".","7","9"]
+]
+输出: true
+```
+
+示例 2:
+
+
+```
+输入:
+[
+  ["8","3",".",".","7",".",".",".","."],
+  ["6",".",".","1","9","5",".",".","."],
+  [".","9","8",".",".",".",".","6","."],
+  ["8",".",".",".","6",".",".",".","3"],
+  ["4",".",".","8",".","3",".",".","1"],
+  ["7",".",".",".","2",".",".",".","6"],
+  [".","6",".",".",".",".","2","8","."],
+  [".",".",".","4","1","9",".",".","5"],
+  [".",".",".",".","8",".",".","7","9"]
+]
+输出: false
+解释: 除了第一行的第一个数字从 5 改为 8 以外，空格内其他数字均与 示例1 相同。
+     但由于位于左上角的 3x3 宫内有两个 8 存在, 因此这个数独是无效的。
+```
+
+示例 3:
+
+
+```
+输入: [1,3,5,6], 7
+输出: 4
+```
+
+
+示例 4:
+
+
+```
+输入: [1,3,5,6], 0
+输出: 0
+```
+
+
+
+## 思路解析
+
+### 一次遍历法
+
+#### 思路
+
+这道题因为需要判断数值是否存在，所以用Hash Map是一个很好的选择。
+因为每一行、每一列、每一格都是需要单独进行判断的，所以需要建立三个长度为9的HashMap数组，分别存放行、列、格的数值。
+
+通过一个二层循环遍历这个9*9的数组,把当前格的数值存放到对应的HashMap中，判断之前是否已经存放过了，如果已经存放过那就退出，返回false，如果是.的话那就跳过，这样只需要遍历一边就可以了。
+
+###  动画理解
+
+![](../Animation/HashMap.gif)
+
+#### 代码实现
+
+
+```java
+//时间复杂度：O(n)
+//空间复杂度：O(1)
+class Solution {
+    public boolean isValidSudoku(char[][] board) {
+        HashMap[] row = new HashMap[9];
+        HashMap[] column = new HashMap[9];
+        HashMap[] box = new HashMap[9];
+        for (int i = 0; i < 9; i++) {
+            row[i] = new HashMap(9);
+            column[i] = new HashMap(9);
+            box[i] = new HashMap(9);
+        }
+        for (int i = 0; i < 9; i++) {
+            for (int j = 0; j < 9; j++) {
+                if (board[i][j] == '.') {
+                    continue;
+                }
+                int boxIndex=i / 3 * 3 + j / 3;
+                if ((boolean) row[i].getOrDefault(board[i][j], true)) {
+                    return false;
+                }
+                if ((boolean) column[j].getOrDefault(board[i][j], true)) {
+                    return false;
+                }
+                if ((boolean) box[boxIndex].getOrDefault(board[i][j], true)) {
+                    return false;
+                }
+                row[i].put(board[i][j], false);
+                column[j].put(board[i][j], false);
+                box[boxIndex].put(board[i][j], false);
+            }
+        }
+
+        return true;
+    }
+}
+```
+
+ ![](../../Pictures/qrcode.jpg)

0036-valid-sudoku/Code/1.java

@@ -0,0 +1,34 @@
+class Solution {
+    public boolean isValidSudoku(char[][] board) {
+        HashMap[] row = new HashMap[9];
+        HashMap[] column = new HashMap[9];
+        HashMap[] box = new HashMap[9];
+        for (int i = 0; i < 9; i++) {
+            row[i] = new HashMap(9);
+            column[i] = new HashMap(9);
+            box[i] = new HashMap(9);
+        }
+        for (int i = 0; i < 9; i++) {
+            for (int j = 0; j < 9; j++) {
+                if (board[i][j] == '.') {
+                    continue;
+                }
+                int boxIndex=i / 3 * 3 + j / 3;
+                if ((boolean) row[i].getOrDefault(board[i][j], true)) {
+                    return false;
+                }
+                if ((boolean) column[j].getOrDefault(board[i][j], true)) {
+                    return false;
+                }
+                if ((boolean) box[boxIndex].getOrDefault(board[i][j], true)) {
+                    return false;
+                }
+                row[i].put(board[i][j], false);
+                column[j].put(board[i][j], false);
+                box[boxIndex].put(board[i][j], false);
+            }
+        }
+
+        return true;
+	}
+}