300. Longest Increasing Subsequence

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Longest Increasing Subsequence

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Given an unsorted array of integers, find the length of longest increasing subsequence.
Example:

Input: [10,9,2,5,3,7,101,18]
Output: 4
Explanation: The longest increasing subsequence is [2,3,7,101], therefore the length is 4.
Note:
	* 
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?

递归图解LIS

递推图解

解法一

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public class LongestIncreasingSubsequence {

   /**
    * dp O(n^2)
    *
    * @param nums
    * @return
    */
   public int lengthOfLIS(int[] nums) {
      if (nums == null || nums.length == 0)
         return 0;
      // 记录每次的结果
      int[] dp = new int[nums.length];
      int ans = 0;
      for (int i = 0; i < nums.length; i++) {
         dp[i] = 1;
         for (int j = 0; j < i; j++) {
            if (nums[j] < nums[i]) {
               dp[i] = dp[i] > dp[j] + 1 ? dp[i] : dp[j] + 1;
            }
         }
         if (dp[i] > ans) {
            ans = dp[i];
         }
      }
      return ans;
   }

解法二:

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  /**
    *
    * O(nlogn)
    *
    * @param nums
    * @return
    */
   public int lengthOfLIS2(int[] nums) {
      if (nums == null || nums.length == 0)
         return 0;
      int[] minLast = new int[nums.length + 1];
      minLast[0] = Integer.MIN_VALUE;
      for (int i = 1; i <= nums.length; i++) {
         minLast[i] = Integer.MAX_VALUE;
      }

      for (int i = 0; i < nums.length; i++) {
         // find the first number in minLast >= nums[i]
         int index = binarySearch(minLast, nums[i]);
         minLast[index] = nums[i];
      }

      for (int i = nums.length; i >= 1; i--) {
         if (minLast[i] != Integer.MAX_VALUE)
            return i;
      }

      return 0;
   }

   private int binarySearch(int[] minLast, int num) {
      int start = 0, end = minLast.length - 1;
      while (start + 1 < end) {
         int mid = (end - start) / 2 + start;
         if (minLast[mid] < num) {
            start = mid;
         } else {
            end = mid;
         }
      }
      return end;
   }

}