26. Remove Duplicates from Sorted Array

coderbean · coderbean · commit 4f7fb97068bd · 2019-03-08T20:04:57.000+08:00
diff --git a/src/leetcode/_26_/Main.java b/src/leetcode/_26_/Main.java
@@ -0,0 +1,19 @@
+package leetcode._26_;
+
+/**
+ * Created by zhangbo54 on 2019-03-04.
+ */
+public class Main {
+    public static void main(String[] args) {
+        Solution solution = new Solution();
+        int[] a = new int[]{1, 2, 2};
+        System.out.println(solution.removeDuplicates(a));
+        System.out.println(a);
+        // int[] a = new int[]{0, 0, 1, 1, 1, 2, 2, 3, 3, 4};
+        // System.out.println(solution.removeDuplicates(a));
+        // for (int i : a) {
+        //     System.out.println(i);
+        // }
+    }
+}
+
diff --git a/src/leetcode/_26_/Solution.java b/src/leetcode/_26_/Solution.java
@@ -0,0 +1,26 @@
+package leetcode._26_;
+
+class Solution {
+    public int removeDuplicates(int[] nums) {
+        if (nums.length == 0) {
+            return 0;
+        }
+        int len = 1;
+        int emptyIndex = -1; // 可以放下一个元素的位置
+        for (int i = 0; i < nums.length - 1; i++) {
+            if (nums[i] == nums[i + 1] && emptyIndex == -1) {
+                emptyIndex = i + 1;
+            }
+            // if (i + 2 == nums.length && emptyIndex == -1){ // 表示全部不重复
+            //     return nums.length;
+            // }
+            if (nums[i] != nums[i + 1]) {
+                len++;
+                if (emptyIndex!=-1) {
+                    nums[emptyIndex++] = nums[i + 1];
+                }
+            }
+        }
+        return len;
+    }
+}
diff --git a/src/leetcode/_26_/solution.md b/src/leetcode/_26_/solution.md
@@ -0,0 +1,75 @@
+### [26\. Remove Duplicates from Sorted Array](https://leetcode.com/problems/remove-duplicates-from-sorted-array/)
+
+Difficulty: **Easy**
+
+
+Given a sorted array _nums_, remove the duplicates such that each element appear only _once_ and return the new length.
+
+Do not allocate extra space for another array, you must do this by **modifying the input array** with O(1) extra memory.
+
+**Example 1:**
+
+```
+Given nums = [1,1,2],
+
+Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively.
+
+It doesn't matter what you leave beyond the returned length.```
+
+**Example 2:**
+
+```
+Given nums = [0,0,1,1,1,2,2,3,3,4],
+
+Your function should return length = 5, with the first five elements of nums being modified to 0, 1, 2, 3, and 4 respectively.
+
+It doesn't matter what values are set beyond the returned length.
+```
+
+**Clarification:**
+
+Confused why the returned value is an integer but your answer is an array?
+
+Note that the input array is passed in by **reference**, which means modification to the input array will be known to the caller as well.
+
+Internally you can think of this:
+
+```
+// nums is passed in by reference. (i.e., without making a copy)
+int len = removeDuplicates(nums);
+
+// any modification to nums in your function would be known by the caller.
+// using the length returned by your function, it prints the first len elements.
+for (int i = 0; i < len; i++) {
+    print(nums[i]);
+}```
+
+
+#### Solution
+
+Language: **Java**
+
+```java
+class Solution {
+    public int removeDuplicates(int[] nums) {
+        if (nums.length == 0) {
+            return 0;
+        }
+        int len = 1;
+        int emptyIndex = -1; // 可以放下一个元素的位置
+        for (int i = 0; i < nums.length - 1; i++) {
+            if (nums[i] == nums[i + 1] && emptyIndex == -1) {
+                emptyIndex = i + 1;
+            }
+            if (nums[i] != nums[i + 1]) {
+                len++;
+                if (emptyIndex!=-1) {
+                    nums[emptyIndex++] = nums[i + 1];
+                }
+            }
+        }
+        return len;
+    }
+}
+```
+![](https://ws3.sinaimg.cn/large/006tKfTcgy1g0vn3pe58aj30vj0u0af2.jpg)
