add LongestIncreasingSubsequence

company/microsoft/LongestIncreasingSubsequence.java

@@ -0,0 +1,34 @@
+//Given an unsorted array of integers, find the length of longest increasing subsequence.
+
+//For example,
+//Given [10, 9, 2, 5, 3, 7, 101, 18],
+//The longest increasing subsequence is [2, 3, 7, 101], therefore the length is 4. Note that there may be more than one LIS combination, it is only necessary for you to return the length.
+
+//Your algorithm should run in O(n2) complexity.
+
+//Follow up: Could you improve it to O(n log n) time complexity?
+
+class LongestIncreasingSubsequence {
+    public int lengthOfLIS(int[] nums) {
+        if(nums == null || nums.length < 1) {
+            return 0;
+        }
+
+        int[] dp = new int[nums.length];
+        dp[0] = 1;
+
+        int max = 1;
+        for(int i = 1; i < dp.length; i++) {
+            int currentMax = 0;
+            for(int j = 0; j < i; j++) {
+                if(nums[i] > nums[j]) {
+                    currentMax = Math.max(currentMax, dp[j]);
+                }
+            }
+            dp[i] = 1 + currentMax;
+            max = Math.max(max, dp[i]);
+        }
+
+        return max;
+    }
+}

leetcode/dynamic-programming/LongestIncreasingSubsequence.java

@@ -0,0 +1,34 @@
+//Given an unsorted array of integers, find the length of longest increasing subsequence.
+
+//For example,
+//Given [10, 9, 2, 5, 3, 7, 101, 18],
+//The longest increasing subsequence is [2, 3, 7, 101], therefore the length is 4. Note that there may be more than one LIS combination, it is only necessary for you to return the length.
+
+//Your algorithm should run in O(n2) complexity.
+
+//Follow up: Could you improve it to O(n log n) time complexity?
+
+class LongestIncreasingSubsequence {
+    public int lengthOfLIS(int[] nums) {
+        if(nums == null || nums.length < 1) {
+            return 0;
+        }
+
+        int[] dp = new int[nums.length];
+        dp[0] = 1;
+
+        int max = 1;
+        for(int i = 1; i < dp.length; i++) {
+            int currentMax = 0;
+            for(int j = 0; j < i; j++) {
+                if(nums[i] > nums[j]) {
+                    currentMax = Math.max(currentMax, dp[j]);
+                }
+            }
+            dp[i] = 1 + currentMax;
+            max = Math.max(max, dp[i]);
+        }
+
+        return max;
+    }
+}