674. Longest Continuous Increasing Subsequence。

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.

给定一个数组，其中需要找到数组中连续递增的最大长度。

第一种可以比较相邻的两个数字的大小关系，如果当前的比前一个的大，就让记录递增连续值的变量加一，如果当前的比前一个的小的话，则让变量重置为1。中间记录下最大的递增连续值最后返回即可。

class Solution {
public:
int findLengthOfLCIS(vector<int>& nums) {

int n = nums.size();

if(n <= 1) {
return n;
}

int maxNum = 1;//记录出现最大连续值
int count = 1;//记录当前连续出现的次数

for(int i=1; i < n; i++) {
//cout << nums[i-1] << "," << nums[i] << endl;
if(nums[i-1] < nums[i]) {//如果比当前的小就是递增
count++;
} else {
count = 1;
}
maxNum = max(count,maxNum);
}
return maxNum;
}
};

第二种可以利用下标位置来计算，需要使用一个变量来记录递增序列开始时的下标，然后根据当前的下标计算出递增序列的长度，然后与最大的递增连续次数进行对比替换即可。

class Solution {
public:
int findLengthOfLCIS(vector<int>& nums) {
int maxLen = 0;//最大的连续次数
int curIndex = 0;//当前的下标

for(int i=0;i<nums.size();i++) {
if(i>0 && nums[i] <= nums[i-1]) {//从第二个开始，记录下递增序列开始的位置
curIndex = i;//记录当前的下标
}
maxLen = max(maxLen,i-curIndex+1);//用当前的最长连续次数和当前的连续次数
}
return maxLen;
}
};