refactor 349

fishercoder1534 · fishercoder1534 · commit d8a50e0d4c15 · 2019-01-20T08:08:23.000-08:00
diff --git a/src/main/java/com/fishercoder/solutions/_349.java b/src/main/java/com/fishercoder/solutions/_349.java
@@ -5,146 +5,78 @@
 import java.util.Iterator;
 import java.util.Set;
 
-/**349. Intersection of Two Arrays
+/**
+ * 349. Intersection of Two Arrays
  *
-Given two arrays, write a function to compute their intersection.
-
-Example:
-Given nums1 = [1, 2, 2, 1], nums2 = [2, 2], return [2].
-
-Note:
-Each element in the result must be unique.
-The result can be in any order.*/
+ * Given two arrays, write a function to compute their intersection.
+ *
+ * Example: Given nums1 = [1, 2, 2, 1], nums2 = [2, 2], return [2].
+ *
+ * Note: Each element in the result must be unique. The result can be in any order.
+ */
 public class _349 {
 
-	//then I clicked its Tags, and find it's marked with so many tags: Binary Search, HashTable, Two Pointers, Sort, now I'll try to do it one by one
-	//inspired by this post: https://discuss.leetcode.com/topic/45685/three-java-solutions
-	public int[] intersection_two_pointers(int[] nums1, int[] nums2) {
-		Set<Integer> set = new HashSet();
-		Arrays.sort(nums1);
-		Arrays.sort(nums2);
-		int i = 0;
-		int j = 0;
-		for (; i < nums1.length && j < nums2.length; ) {
-			if (nums1[i] < nums2[j]) {
-				i++;
-			} else if (nums1[i] > nums2[j]) {
-				j++;
-			} else {
-				set.add(nums1[i]);
-				i++;
-				j++;
-			}
-		}
-		int[] result = new int[set.size()];
-		Iterator<Integer> it = set.iterator();
-		int k = 0;
-		while (it.hasNext()) {
-			result[k++] = it.next();
-		}
-		return result;
-	}
-
-	public int[] intersection_binary_search(int[] nums1, int[] nums2) {
-		//this approach is O(nlgn)
-		Arrays.sort(nums1);
-		Arrays.sort(nums2);
-		Set<Integer> intersect = new HashSet();
-		for (int i : nums1) {
-			if (binarySearch(i, nums2)) {
-				intersect.add(i);
-			}
-		}
-		int[] result = new int[intersect.size()];
-		Iterator<Integer> it = intersect.iterator();
-		for (int i = 0; i < intersect.size(); i++) {
-			result[i] = it.next();
-		}
-		return result;
-	}
-
-	private boolean binarySearch(int i, int[] nums) {
-		int left = 0;
-		int right = nums.length - 1;
-		while (left <= right) {
-			int mid = left + (right - left) / 2;
-			if (nums[mid] == i) {
-				return true;
-			} else if (nums[mid] > i) {
-				right = mid - 1;
-			} else {
-				left = mid + 1;
-			}
-		}
-		return false;
-	}
-
-	//tried a friend's recommended approach, didn't finish it to get it AC'ed, turned to normal approach as above and got it AC'ed.
-	private boolean binarySearch_not_working_version(int i, int[] nums) {
-		if (nums == null || nums.length == 0) {
-			return false;
-		}
-		int left = 0;
-		int right = nums.length - 1;
-		while (left + 1 < right) {
-			int mid = left + (right - left) / 2;
-			if (nums[mid] > i) {
-				right = mid;
-			} else if (nums[mid] < 1) {
-				left = mid;
-			} else if (nums[mid] == i) {
-				return true;
-			} else {
-				return false;
-			}
-		}
-		return nums[left] == i || nums[right] == i;
-	}
-
-	public static void main(String... strings) {
-		_349 test = new _349();
-		int[] nums1 = new int[]{1, 2};
-		int[] nums2 = new int[]{2, 1};
-		test.intersection_binary_search(nums1, nums2);
-	}
-
-	public int[] intersection_two_hashsets(int[] nums1, int[] nums2) {
-		//this approach is O(n)
-		Set<Integer> set1 = new HashSet();
-		for (int i = 0; i < nums1.length; i++) {
-			set1.add(nums1[i]);
-		}
-		Set<Integer> intersect = new HashSet();
-		for (int i = 0; i < nums2.length; i++) {
-			if (set1.contains(nums2[i])) {
-				intersect.add(nums2[i]);
-			}
-		}
-		int[] result = new int[intersect.size()];
-		Iterator<Integer> it = intersect.iterator();
-		for (int i = 0; i < intersect.size(); i++) {
-			result[i] = it.next();
-		}
-		return result;
-	}
+  public static class Solution1 {
+    public int[] intersection(int[] nums1, int[] nums2) {
+      Set<Integer> set = new HashSet();
+      Arrays.sort(nums1);
+      Arrays.sort(nums2);
+      int i = 0;
+      int j = 0;
+      for (; i < nums1.length && j < nums2.length; ) {
+        if (nums1[i] < nums2[j]) {
+          i++;
+        } else if (nums1[i] > nums2[j]) {
+          j++;
+        } else {
+          set.add(nums1[i]);
+          i++;
+          j++;
+        }
+      }
+      int[] result = new int[set.size()];
+      Iterator<Integer> it = set.iterator();
+      int k = 0;
+      while (it.hasNext()) {
+        result[k++] = it.next();
+      }
+      return result;
+    }
+  }
 
-	//so naturally, I come up with this naive O(n^2) solution and surprisingly it got AC'ed immediately, no wonder it's marked as EASY.
-	public int[] intersection_naive(int[] nums1, int[] nums2) {
-		Set<Integer> set = new HashSet();
-		for (int i = 0; i < nums1.length; i++) {
-			for (int j = 0; j < nums2.length; j++) {
-				if (nums1[i] == nums2[j]) {
-					set.add(nums1[i]);
-				}
-			}
-		}
-		int[] result = new int[set.size()];
-		Iterator<Integer> it = set.iterator();
-		int i = 0;
-		while (it.hasNext()) {
-			result[i++] = it.next();
-		}
-		return result;
-	}
+  public static class Solution2 {
+    public int[] intersection(int[] nums1, int[] nums2) {
+      //this approach is O(nlgn)
+      Arrays.sort(nums1);
+      Arrays.sort(nums2);
+      Set<Integer> intersect = new HashSet();
+      for (int i : nums1) {
+        if (binarySearch(i, nums2)) {
+          intersect.add(i);
+        }
+      }
+      int[] result = new int[intersect.size()];
+      Iterator<Integer> it = intersect.iterator();
+      for (int i = 0; i < intersect.size(); i++) {
+        result[i] = it.next();
+      }
+      return result;
+    }
 
+    private boolean binarySearch(int i, int[] nums) {
+      int left = 0;
+      int right = nums.length - 1;
+      while (left <= right) {
+        int mid = left + (right - left) / 2;
+        if (nums[mid] == i) {
+          return true;
+        } else if (nums[mid] > i) {
+          right = mid - 1;
+        } else {
+          left = mid + 1;
+        }
+      }
+      return false;
+    }
+  }
 }
