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.
这题挺好的，我想到两种方法，一种brute force，O(n2)不好；另一种，看到LIS就思维定势地想到了DP。
DP:
我写的leetcode submissions没保存，摘抄一个别人的：
public int findLengthOfLCIS(int[] nums) {
if (nums == null || nums.length == 0) return 0;
int n = nums.length;
int[] dp = new int[n];
int max = 1;
dp[0] = 1;
for (int i = 1; i < n; i++) {
if (nums[i] > nums[i - 1]) {
dp[i] = dp[i - 1] + 1;
}
else {
dp[i] = 1;
}
max = Math.max(max, dp[i]);
}
return max;
}
但其实这题有更好的方法，
public int findLengthOfLCIS(int[] nums) {
int res = 0, cnt = 0;
for(int i = 0; i < nums.length; i++){
if(i == 0 || nums[i-1] < nums[i]) res = Math.max(res, ++cnt);
else cnt = 1;
}
return res;
}
用O(1)记录某个值然后随着滚动而重置，这种思想。
