Question

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

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.

Difficulty:Easy

Category:Array, DP

Analyze

Solution

Solution 1: 动态规划求解

class Solution {
 public:
  int findLengthOfLCIS(vector<int>& nums) {
    if (nums.empty()) return 0;
    vector<int> res(nums.size(), 1);
    for (int i = 1; i < nums.size(); ++i) {
      if (nums[i] > nums[i - 1]) res[i] = res[i - 1] + 1;
    }

    int max_res = 1;
    for (int i : res) max_res = max(max_res, i);
    return max_res;
  }
};

Solution 2: 优化

class Solution {
 public:
  int findLengthOfLCIS(vector<int>& nums) {
    if (nums.empty()) return 0;
    int cur = 1;
    int ans = 1;

    for (int i = 1; i < nums.size(); ++i) {
      if (nums[i] > nums[i - 1]) {
        ++cur;
        ans = max(cur, ans);
      } else
        cur = 1;
    }

    return ans;
  }
};