CrackerCat
diff --git a/‎note/015/README.md
Copy file name to clipboardExpand all lines: note/015/README.md
+16-9Lines changed: 16 additions & 9 deletions b/‎note/015/README.md
Copy file name to clipboardExpand all lines: note/015/README.md
+16-9Lines changed: 16 additions & 9 deletions
diff --git a/‎src/com/blankj/medium/_015/Solution.java
Copy file name to clipboardExpand all lines: src/com/blankj/medium/_015/Solution.java
+15-8Lines changed: 15 additions & 8 deletions b/‎src/com/blankj/medium/_015/Solution.java
Copy file name to clipboardExpand all lines: src/com/blankj/medium/_015/Solution.java
+15-8Lines changed: 15 additions & 8 deletions
@@ -21,27 +21,34 @@ A solution set is:
 
 ## 思路
 
-题意是让你从数组中找出所有三个和为 0 的元素构成的非重复序列，这样的话我们可以把数组先做下排序，然后遍历这个排序数组，用两个指针分别指向当前元素的下一个和数组尾部，判断三者的和与 0 的大小来移动两个指针，其中细节操作就是注意去重。
+题意是让你从数组中找出所有三个和为 0 的元素构成的非重复序列，这样的话我们可以把数组先做下排序，然后遍历这个排序数组，用两个指针分别指向当前元素的下一个和数组尾部，判断三者的和与 0 的大小来移动两个指针，其中细节操作就是优化和去重。
 
 ```java
 class Solution {
     public List<List<Integer>> threeSum(int[] nums) {
-        Arrays.sort(nums);
         List<List<Integer>> list = new ArrayList<>();
-        int i = 0;
-        while (i < nums.length - 2) {
+        int len = nums.length;
+        if (len < 3) return list;
+        Arrays.sort(nums);
+        int max = nums[len - 1];
+        if (max < 0) return list;
+        for (int i = 0; i < len - 2; ) {
             if (nums[i] > 0) break;
-            int left = i + 1, right = nums.length - 1;
+            if (nums[i] + 2 * max < 0) {
+                while (nums[i] == nums[++i] && i < len - 2) ;
+                continue;
+            }
+            int left = i + 1, right = len - 1;
             while (left < right) {
                 int sum = nums[i] + nums[left] + nums[right];
                 if (sum == 0) {
-                    list.add(Arrays.asList(nums[i], nums[left++], nums[right--]));
-                    while (left < right && nums[left] == nums[left - 1]) ++left;
-                    while (left < right && nums[right] == nums[right + 1]) --right;
+                    list.add(Arrays.asList(nums[i], nums[left], nums[right]));
+                    while (nums[left] == nums[++left] && left < right) ;
+                    while (nums[right] == nums[--right] && left < right) ;
                 } else if (sum < 0) ++left;
                 else --right;
             }
-            while (nums[i] == nums[++i] && i < nums.length - 2) ;
+            while (nums[i] == nums[++i] && i < len - 2) ;
         }
         return list;
     }
 
@@ -14,22 +14,29 @@
  */
 public class Solution {
     public List<List<Integer>> threeSum(int[] nums) {
-        Arrays.sort(nums);
         List<List<Integer>> list = new ArrayList<>();
-        int i = 0;
-        while (i < nums.length - 2) {
+        int len = nums.length;
+        if (len < 3) return list;
+        Arrays.sort(nums);
+        int max = nums[len - 1];
+        if (max < 0) return list;
+        for (int i = 0; i < len - 2; ) {
             if (nums[i] > 0) break;
-            int left = i + 1, right = nums.length - 1;
+            if (nums[i] + 2 * max < 0) {
+                while (nums[i] == nums[++i] && i < len - 2) ;
+                continue;
+            }
+            int left = i + 1, right = len - 1;
             while (left < right) {
                 int sum = nums[i] + nums[left] + nums[right];
                 if (sum == 0) {
-                    list.add(Arrays.asList(nums[i], nums[left++], nums[right--]));
-                    while (left < right && nums[left] == nums[left - 1]) ++left;
-                    while (left < right && nums[right] == nums[right + 1]) --right;
+                    list.add(Arrays.asList(nums[i], nums[left], nums[right]));
+                    while (nums[left] == nums[++left] && left < right) ;
+                    while (nums[right] == nums[--right] && left < right) ;
                 } else if (sum < 0) ++left;
                 else --right;
             }
-            while (nums[i] == nums[++i] && i < nums.length - 2) ;
+            while (nums[i] == nums[++i] && i < len - 2) ;
         }
         return list;
     }