674. Longest consecutive increasing sequence
Given an
未经排序integer array, find the longest and连续increasing sequence, and return that sequence长度.
Example 1:
输入: [1,3,5,4,7]
输出: 3
解释: 最长连续递增序列是 [1,3,5], 长度为3。
尽管 [1,3,5,7] 也是升序的子序列, 但它不是连续的，因为5和7在原数组里被4隔开。
Example 2:
输入: [2,2,2,2,2]
输出: 1
解释: 最长连续递增序列是 [2], 长度为1。
Note: The length of the array will not exceed 10,000.
Ideas: continuous up, so only consider nums[i]and its front nums[i-1]can
Solution one :
class Solution {
public int findLengthOfLCIS(int[] nums) {
int LEN = nums.length;
if(LEN == 0 || nums == null) return 0;
int[] dp = new int[LEN];
Arrays.fill(dp, 1);
int res = 0;
for(int i = 1; i < LEN; i++){
if(nums[i-1] < nums[i]){
dp[i] = dp[i-1]+1;
}
res = Math.max(res, dp[i]);
}
return (res > 1)? res:1;
}
}
Time complexity: O(n)
Space compression: O(n)
Solution 2 : Sliding window (state compression), only need to save the previous state, not all states
class Solution {
public int findLengthOfLCIS(int[] nums) {
int LEN = nums.length;
if(LEN == 0 || nums == null) return 0;
int pre = 1;
int cur = 1;
int res = 0;
for(int i = 1; i < LEN; i++){
pre = cur;
if(nums[i-1] < nums[i]){
cur = pre + 1;
}else{
cur = 1;
}
res = Math.max(res, cur);
}
return (res > 1)? res:1;
}
}
Time complexity: O(n)
Space compression: O(1)