Merge pull request #137 from qilingit/day26

Add Java solution for day 26: Contiguous Array

Author: jinshendan

May-LeetCoding-Challenge/26-Contiguous-Array/Contiguous-Array.java

@@ -0,0 +1,42 @@
+/**
+ * Use a hashMap to stock the index and the count, the count minus 1 if reach 0, plus 1 if reach 1.
+ * We compare current length of contiguous array with the maxLength
+ */
+/**
+ * Time complexity: O(n), because a signle loop of nums[]
+ * Space complexity: O(n), becasue we build a HashMap
+ */
+class Solution {
+    public int findMaxLength(final int[] nums) {
+        int maxLength = 0;
+        int count = 0;
+        final Map<Integer, Integer> countsMap = new HashMap<>();
+        
+        for (int i = 0; i < nums.length; i++) {
+            if (nums[i] == 0) {
+                count -= 1;
+            } else {
+                count += 1;
+            }
+            
+            /**
+             * if the count is 0, it means that from the start of index, we have nothing added to the count，
+             * So the length of contiguous array is from the starter to current index
+             */
+            if (count == 0) {
+                maxLength = Math.max(maxLength, (i + 1));
+            } else if (countsMap.containsKey(count)) {
+                /**
+                 * If the count is contained in the map, it means that for the index of contained count to 
+                 * the current index, we have nothing added, we can build a contignous array, it's length is
+                 * i minus index of contained count
+                 */
+                maxLength = Math.max(maxLength, i - countsMap.get(count));
+            } else {
+                countsMap.put(count, i);
+            }
+            
+        }
+        return maxLength;
+    }
+}