674. Longest Continuous Increasing Subsequence

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

Example 1: Input: [1,3,5,4,7] Output: 3 Explanation: The longest continuous increasing subsequence is [1,3,5], its length is 3. Even though [1,3,5,7] is also an increasing subsequence, it's not a continuous one where 5 and 7 are separated by 4. Example 2: Input: [2,2,2,2,2] Output: 1 Explanation: The longest continuous increasing subsequence is [2], its length is 1. Note: Length of the array will not exceed 10,000.

/ Intuition: 计算相邻的差，如果大于0，前进直至遇到小于0，记录长度，以此类推，遍历完数组，len = max(len, i)； /

public int findLengthOfLCIS(int[] nums) {
   //构造数组sum，用于存放数组nums的差值
    int []sum = new int[nums.length+1];
    int []max_num = new int[nums.length];
    for (int i = 1; i < nums.length; i++) {
        sum[i] = nums[i]  - nums[i-1];
    }
    //对数组sum操作
    int len = 0, j = 0, status=0;
    for (int i = 1; i < nums.length; i++) {
        if(sum[i] > 0)
            len++;
        else {
            max_num[++j] = len;
            len = 0;
        }       
    }

    //遍历找出最大子序列长度
    for (int i = 0; i < max_num.length; i++) {
        len = Math.max(max_num[i],len);
    }
    if(nums.length <= 0)
        len = -1;
    return len+1;
}    



//上面代码空间占用太多，删掉数组sum和max_num，直接在nums操作

public static int findLengthOfLCIS(int[] nums) {
 //操作数组nums，将相邻的差值存储在i-1
     for (int i = 1; i < nums.length; i++) {
      nums[i-1] = nums[i]  - nums[i-1];
   }

     int max_len=0,len = 0;
     for (int i = 0; i < nums.length-1; i++) {
      if(nums[i] > 0)
         len++;
      else {
         max_len = Math.max(max_len, len);
         len = 0;
      }     
   }

     max_len = Math.max(max_len, len);
     if(nums.length <= 0)
      max_len = -1;

     return max_len+1;
}