MEDIUM/src/medium/_4Sum.java

fishercoder1534 · fishercoder1534 · commit bb55054e975b · 2016-09-08T13:03:48.000-07:00
diff --git a/MEDIUM/src/medium/_4Sum.java b/MEDIUM/src/medium/_4Sum.java
@@ -0,0 +1,41 @@
+package medium;
+
+import java.util.ArrayList;
+import java.util.Arrays;
+import java.util.List;
+
+public class _4Sum {
+    /**Then I went online and found that there're really smarter/more efficient solutions out there:
+     * Post 1: https://discuss.leetcode.com/topic/29585/7ms-java-code-win-over-100
+     *  this post basically uses two subfunctions, it keeps using binary search to divide this 4sum to 3sum and then to 2sum,
+     *  thus, it's really fast!!!! Instead of using two for loops.*/
+    
+/**My original solution:*/
+
+
+    public List<List<Integer>> fourSum(int[] nums, int target) {
+        List<List<Integer>> result = new ArrayList();
+        if(nums == null || nums.length == 0) return result;
+        Arrays.sort(nums);
+        for(int i = 0; i < nums.length -3; i++){
+            if(i > 0 && nums[i-1] == nums[i]) continue;
+            for(int j = i+1; j < nums.length-2; j++){
+                if(j > i+1 && nums[j-1] == nums[j]) continue;
+                int left = j+1, right = nums.length-1;
+                while(left < right){
+                    int sum = nums[i] + nums[j] + nums[left] + nums[right];
+                    if(sum == target){
+                        result.add(Arrays.asList(nums[i], nums[j], nums[left], nums[right]));
+                        while(left+1 < right && nums[left] == nums[left+1]) left++;
+                        while(right-1 > left && nums[right] == nums[right-1]) right--;
+                        left++;
+                        right--;
+                    } else if(sum > target) right--;
+                    else left++;
+                }
+            }
+        }
+        return result;
+    }
+
+}
