nine chapter ch02 binary search and sorted array

anduo · anduo · commit f65144ac34a5 · 2017-03-13T20:36:53.000+08:00
diff --git a/readme.md b/readme.md
@@ -187,4 +187,38 @@ doc中有好几份文档，比较经典，可以参考。
 ## chapter02 binary search & sorted array
 这一部分都可以使用bs模板来解决问题 [二分搜索模板](./src/template/bs_template.md)
 
+算法面试中如果需要优化O(n)的时间复杂度，那么只能是O(logn)的二分法
+
+**二分搜索四点要素**:
+1. start + 1 < end
+2. start + (end - start) / 2
+3. A[mid] ==, <, >
+4. A[start] A[end] ? target
+
 - [question01:first position of target](./src/ninechapter/ch02_binary_search_and_sorted_array/Question01FirstPositionOfTarget.java) || [lintcode](http://www.lintcode.com/zh-cn/problem/first-position-of-target/)
+- [question02:search for a range](./src/ninechapter/ch02_binary_search_and_sorted_array/Question02SearchForARange.java)
+- [question03:search insert position](./src/ninechapter/ch02_binary_search_and_sorted_array/Question03SearchInsertPosition.java)
+- [question04:search a 2D matrix](./src/ninechapter/ch02_binary_search_and_sorted_array/Question04SearchA2DMatrix.java)
+- [question05:search a 2D matrix ii](./src/ninechapter/ch02_binary_search_and_sorted_array/Question05SearchA2DMatrixII.java)
+- [question06:first bad version](./src/ninechapter/ch02_binary_search_and_sorted_array/Question06FirstBadVersion.java)
+- [question07:find peak element](./src/ninechapter/ch02_binary_search_and_sorted_array/Question07FindPeakElement.java)
+- [question08:find minimum in rotated sorted array](./src/ninechapter/ch02_binary_search_and_sorted_array/Question08FindMinimumInRotatedSortedArray.java)
+- [question09:find minimum in rotated sorted array ii](./src/ninechapter/ch02_binary_search_and_sorted_array/Question08FindMinimumInRotatedSortedArrayII.java)
+- [question10:search in rotated sorted array](./src/ninechapter/ch02_binary_search_and_sorted_array/Question10SearchInRotatedSortedArray.java)
+- [question11:search in rotated sorted array ii](./src/ninechapter/ch02_binary_search_and_sorted_array/Question10SearchInRotatedSortedArrayII.java)
+- [question12:median of two sorted arrays](./src/ninechapter/ch02_binary_search_and_sorted_array/Question12MedianOfTwoSortedArrays.java)
+- [question13:recover rotated sorted array](./src/ninechapter/ch02_binary_search_and_sorted_array/Question13RecoverRotatedSortedArray.java)
+- [question14:rotate string](./src/ninechapter/ch02_binary_search_and_sorted_array/Question14RotateString.java)
+- [question15:reverse words in a string](./src/ninechapter/ch02_binary_search_and_sorted_array/Question15ReverseWordsInAString.java)
+
+**总结**
+1. Binary Search Template (4 key points)
+2. Rotated Sorted Array
+    1. Find Minimum
+    2. Find Target
+    3. why o(n) with duplicates ?
+3. Find Median in Two Sorted Array
+find kth
+4. Reverse in 3 steps
+
+
diff --git a/src/ninechapter/ch02_binary_search_and_sorted_array/Question04SearchA2DMatrix.java b/src/ninechapter/ch02_binary_search_and_sorted_array/Question04SearchA2DMatrix.java
@@ -0,0 +1,76 @@
+package ninechapter.ch02_binary_search_and_sorted_array;
+/**
+ * 写出一个高效的算法来搜索 m × n矩阵中的值。
+ * 这个矩阵具有以下特性：
+ * 每行中的整数从左到右是排序的。
+ * 每行的第一个数大于上一行的最后一个整数。
+ * <p>
+ * [
+ * [1, 3, 5, 7],
+ * [10, 11, 16, 20],
+ * [23, 30, 34, 50]
+ * ]
+ * target = 3,返回 true
+ */
+
+/**
+ * Created by anduo on 17-3-13.
+ */
+public class Question04SearchA2DMatrix {
+    public boolean searchMatrix(int[][] matrix, int target) {
+        // 二分
+        if (matrix == null
+                || matrix.length == 0
+                || matrix[0] == null
+                || matrix[0].length == 0) {
+            return false;
+        }
+
+        int rows = matrix.length;
+        int cols = matrix[0].length;
+
+        int row;// 需要找的目标行
+        // 先找到行
+        int start = 0;
+        int end = rows - 1;
+        while (start + 1 < end) {
+            int mid = start + (end - start) / 2;
+            int v = matrix[mid][0];
+            if (v == target) {
+                return true;
+            } else if (v > target) {// 说明一定在mid行的上班部分
+                end = mid;
+            } else {// 在mid行的下半部分
+                start = mid;
+            }
+        }
+        if (matrix[end][0] <= target) {// 如果在上半部分
+            row = end;
+        } else if (matrix[start][0] <= target) {
+            row = start;
+        } else {
+            return false;
+        }
+
+        // 找列
+        start = 0;
+        end = cols - 1;
+        while (start + 1 < end) {
+            int mid = start + (end - start) / 2;
+            int v = matrix[row][mid];
+            if (v == target) {
+                return true;
+            } else if (v > target) {// 说明一定在mid列的前半部分
+                end = mid;
+            } else {// 在mid列的后部分
+                start = mid;
+            }
+        }
+        if (matrix[row][start] == target) {
+            return true;
+        } else if (matrix[row][end] == target) {
+            return true;
+        }
+        return false;
+    }
+}
diff --git a/src/ninechapter/ch02_binary_search_and_sorted_array/Question05SearchA2DMatrixII.java b/src/ninechapter/ch02_binary_search_and_sorted_array/Question05SearchA2DMatrixII.java
@@ -0,0 +1,50 @@
+package ninechapter.ch02_binary_search_and_sorted_array;
+/**
+ * 写出一个高效的算法来搜索m×n矩阵中的值，返回这个值出现的次数。
+ * 这个矩阵具有以下特性：
+ * 每行中的整数从左到右是排序的。
+ * 每一列的整数从上到下是排序的。
+ * 在每一行或每一列中没有重复的整数。
+ * <p>
+ * [
+ * [1, 3, 5, 7],
+ * [2, 4, 7, 8],
+ * [3, 5, 9, 10]
+ * ]
+ * target = 3,返回 2
+ */
+
+/**
+ * Created by anduo on 17-3-13.
+ */
+public class Question05SearchA2DMatrixII {
+    public int searchMatrix(int[][] matrix, int target) {
+        if (matrix == null
+                || matrix.length == 0
+                || matrix[0] == null
+                || matrix[0].length == 0) {
+            return 0;
+        }
+
+        int rows = matrix.length;
+        int cols = matrix[0].length;
+
+        int x = rows - 1;
+        int y = 0;
+        int count = 0;
+        while (x >= 0 && y < cols) {
+            if (matrix[x][y] == target) {// 如果找到相应的元素
+                count++;
+                x--;
+                y++;
+                // 行列都移动
+            } else if (matrix[x][y] < target) {
+                y++;// 在下一列
+            } else {
+                x--;//在上一行
+            }
+        }
+
+        return count;
+    }
+}
diff --git a/src/ninechapter/ch02_binary_search_and_sorted_array/Question06FirstBadVersion.java b/src/ninechapter/ch02_binary_search_and_sorted_array/Question06FirstBadVersion.java
@@ -0,0 +1,44 @@
+package ninechapter.ch02_binary_search_and_sorted_array;
+
+/**
+ * 代码库的版本号是从 1 到 n 的整数。
+ * 某一天，有人提交了错误版本的代码，因此造成自身及之后版本的代码在单元测试中均出错。
+ * 请找出第一个错误的版本号。
+ * 你可以通过isBadVersion的接口来判断版本号version是否在单元测试中出错，
+ * 具体接口详情和调用方法请见代码的注释部分。
+ * 样例
+ * 给出 n=5
+ * 调用isBadVersion(3)，得到false
+ * 调用isBadVersion(5)，得到true
+ * 调用isBadVersion(4)，得到true
+ * 此时我们可以断定4是第一个错误的版本号
+ * Created by anduo on 17-3-13.
+ */
+public class Question06FirstBadVersion {
+    /**
+     * @param n: An integers.
+     * @return: An integer which is the first bad version.
+     */
+    public int findFirstBadVersion(int n) {
+        int start = 0;
+        int end = n;
+        while (start + 1 < end) {
+            int mid = start + (end - start) / 2;
+            if (SVNRepo.isBadVersion(mid)) {
+                end = mid;
+            } else {
+                start = mid;
+            }
+        }
+        if (SVNRepo.isBadVersion(start)) {
+            return start;
+        }
+        return end;
+    }
+
+    static class SVNRepo {
+        public static boolean isBadVersion(int n) {
+            return true;
+        }
+    }
+}
diff --git a/src/ninechapter/ch02_binary_search_and_sorted_array/Question07FindPeakElement.java b/src/ninechapter/ch02_binary_search_and_sorted_array/Question07FindPeakElement.java
@@ -0,0 +1,44 @@
+package ninechapter.ch02_binary_search_and_sorted_array;
+
+/**
+ * 寻找峰值
+ * http://www.lintcode.com/zh-cn/problem/find-peak-element/
+ * 你给出一个整数数组(size为n)，其具有以下特点：
+ * 相邻位置的数字是不同的
+ * A[0] < A[1] 并且 A[n - 2] > A[n - 1]
+ * 假定P是峰值的位置则满足A[P] > A[P-1]且A[P] > A[P+1]，返回数组中任意一个峰值的位置。
+ * <p>
+ * --------
+ * 给出数组[1, 2, 1, 3, 4, 5, 7, 6]返回1, 即数值 2 所在位置, 或者6, 即数值 7 所在位置.
+ * Created by anduo on 17-3-13.
+ */
+public class Question07FindPeakElement {
+    /**
+     * @param A: An integers array.
+     * @return: return any of peek positions.
+     */
+    public int findPeak(int[] A) {
+        // 局部有序
+        if (A == null || A.length == 0) {
+            return -1;
+        }
+
+        int start = 1;
+        int end = A.length - 2;
+        while (start + 1 < end) {
+            int mid = start + (end - start) / 2;
+            if (A[mid] < A[mid - 1]) {
+                end = mid;
+            } else if (A[mid] < A[mid + 1]) {
+                start = mid;
+            } else {
+                end = mid;
+            }
+        }
+        if (A[start] < A[end]) {
+            return end;
+        } else {
+            return start;
+        }
+    }
+}
diff --git a/src/ninechapter/ch02_binary_search_and_sorted_array/Question08FindMinimumInRotatedSortedArray.java b/src/ninechapter/ch02_binary_search_and_sorted_array/Question08FindMinimumInRotatedSortedArray.java
@@ -0,0 +1,48 @@
+package ninechapter.ch02_binary_search_and_sorted_array;
+
+/**
+ * http://www.lintcode.com/zh-cn/problem/find-minimum-in-rotated-sorted-array/
+ * 假设一个旋转排序的数组其起始位置是未知的
+ * （比如0 1 2 4 5 6 7 可能变成是4 5 6 7 0 1 2）。
+ * 你需要找到其中最小的元素。
+ * 你可以假设数组中不存在重复的元素。
+ * Created by anduo on 17-3-13.
+ */
+public class Question08FindMinimumInRotatedSortedArray {
+    /**
+     * @param nums: a rotated sorted array
+     * @return: the minimum number in the array
+     */
+    public int findMin(int[] nums) {
+        if (nums == null || nums.length == 0) {
+            return -1;
+        }
+        if (nums.length == 1) {
+            return nums[0];
+        }
+
+
+        int start = 0, end = nums.length - 1;
+        int target = nums[end];
+
+        // 4 5 6 7 0 1 2
+        // find the first element <= target
+        while (start + 1 < end) {
+            int mid = start + (end - start) / 2;
+            if (nums[mid] == target) {
+                // if mid equals to end, that means it's fine to remove end
+                // the smallest element won't be removed
+                end--;
+            } else if (nums[mid] <= target) {
+                end = mid;
+            } else {
+                start = mid;
+            }
+        }
+        if (nums[start] <= target) {
+            return nums[start];
+        } else {
+            return nums[end];
+        }
+    }
+}
diff --git a/src/ninechapter/ch02_binary_search_and_sorted_array/Question09FindMinimumInRotatedSortedArrayII.java b/src/ninechapter/ch02_binary_search_and_sorted_array/Question09FindMinimumInRotatedSortedArrayII.java
@@ -0,0 +1,46 @@
+package ninechapter.ch02_binary_search_and_sorted_array;
+
+/**
+ * http://www.lintcode.com/zh-cn/problem/find-minimum-in-rotated-sorted-array-ii/
+ * 假设一个旋转排序的数组其起始位置是未知的
+ * （比如0 1 2 4 5 6 7 可能变成是4 5 6 7 0 1 2）。
+ * 你需要找到其中最小的元素。
+ * 数组中可能存在重复的元素。
+ * Created by anduo on 17-3-13.
+ */
+public class Question09FindMinimumInRotatedSortedArrayII {
+    /**
+     * @param nums: a rotated sorted array
+     * @return: the minimum number in the array
+     */
+    public int findMin(int[] nums) {
+        if (nums == null || nums.length == 0) {
+            return -1;
+        }
+        if (nums.length == 1) {
+            return nums[0];
+        }
+        if (nums.length == 2) {
+            return Math.min(nums[0], nums[1]);
+        }
+
+        int start = 0, end = nums.length - 1;
+        int target = nums[nums.length - 1];
+
+        // 4 5 6 7 0 1 2
+        // find the first element <= target
+        while (start + 1 < end) {
+            int mid = start + (end - start) / 2;
+            if (nums[mid] <= target) {
+                end = mid;
+            } else {
+                start = mid;
+            }
+        }
+        if (nums[start] <= target) {
+            return nums[start];
+        } else {
+            return nums[end];
+        }
+    }
+}
diff --git a/src/ninechapter/ch02_binary_search_and_sorted_array/Question10SearchInRotatedSortedArray.java b/src/ninechapter/ch02_binary_search_and_sorted_array/Question10SearchInRotatedSortedArray.java
@@ -8,7 +8,7 @@
  * 给定一个目标值进行搜索，如果在数组中找到目标值返回数组中的索引位置，否则返回-1。
  * 你可以假设数组中不存在重复的元素。
  */
-public class Question04SearchInRotatedSortedArray {
+public class Question10SearchInRotatedSortedArray {
     public int search(int[] A, int target) {
         if (A == null || A.length == 0) {
             return -1;
diff --git a/src/ninechapter/ch02_binary_search_and_sorted_array/Question11SearchInRotatedSortedArrayII.java b/src/ninechapter/ch02_binary_search_and_sorted_array/Question11SearchInRotatedSortedArrayII.java
@@ -10,7 +10,7 @@
  * 为何会影响？
  * 写出一个函数判断给定的目标值是否出现在数组中。
  */
-public class Question05SearchInRotatedSortedII {
+public class Question11SearchInRotatedSortedArrayII {
     // 这个问题在面试中不会让实现完整程序
     // 只需要举出能够最坏情况的数据是 [1,1,1,1... 1] 里有一个0即可。
     // 在这种情况下是无法使用二分法的，复杂度是O(n)
diff --git a/src/ninechapter/ch02_binary_search_and_sorted_array/Question12MedianOfTwoSortedArrays.java b/src/ninechapter/ch02_binary_search_and_sorted_array/Question12MedianOfTwoSortedArrays.java
diff --git a/src/ninechapter/ch02_binary_search_and_sorted_array/Question13RecoverRotatedSortedArray.java b/src/ninechapter/ch02_binary_search_and_sorted_array/Question13RecoverRotatedSortedArray.java
diff --git a/src/ninechapter/ch02_binary_search_and_sorted_array/Question14RotateString.java b/src/ninechapter/ch02_binary_search_and_sorted_array/Question14RotateString.java
diff --git a/src/ninechapter/ch02_binary_search_and_sorted_array/Question15ReverseWordsInAString.java b/src/ninechapter/ch02_binary_search_and_sorted_array/Question15ReverseWordsInAString.java
