[LeetCode] 674. Longest Continuous Increasing Subsequence 最长连续递增序列

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.

这道题让我们求一个数组的最长连续递增序列，由于有了连续这个条件，跟之前那道 Number of Longest Increasing Subsequence 比起来，其实难度就降低了很多。可以使用一个计数器，如果遇到大的数字，计数器自增1；如果是一个小的数字，则计数器重置为1。用一个变量 cur 来表示前一个数字，初始化为整型最大值，当前遍历到的数字 num 就和 cur 比较就行了，每次用 cnt 来更新结果 res，参见代码如下：

解法一：

class Solution {
public:
    int findLengthOfLCIS(vector<int>& nums) {
        int res = 0, cnt = 0, cur = INT_MAX;
        for (int num : nums) {
            if (num > cur) ++cnt;
            else cnt = 1;
            res = max(res, cnt);
            cur = num;
        }
        return res;
    }
};

下面这种方法的思路和上面的解法一样，每次都和前面一个数字来比较，注意处理无法取到钱一个数字的情况，参见代码如下：

解法二：

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