LeetCode 674. Longest Continuous Increasing Subsequence
2018-02-05 03:35
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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 is 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.
题目大意:给一个乱序数组,返回这个数组中递增连续子序列中最长的长度~
分析:遍历从i
= 1到结尾,每次比较nums[i-1]和nums[i],如果是递增的就将temp++,否则temp=1,每次将temp最大值保存在ans中,最后返回ans的值~
class Solution {
public:
int findLengthOfLCIS(vector<int>& nums) {
int temp = 1, ans = 1;
for (int i = 1; i < nums.size(); i++) {
temp = (nums[i-1] < nums[i]) ? temp + 1 : 1;
ans = max(ans, temp);
}
return nums.size() == 0 ? 0 : ans;
}
};
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 is 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.
题目大意:给一个乱序数组,返回这个数组中递增连续子序列中最长的长度~
分析:遍历从i
= 1到结尾,每次比较nums[i-1]和nums[i],如果是递增的就将temp++,否则temp=1,每次将temp最大值保存在ans中,最后返回ans的值~
class Solution {
public:
int findLengthOfLCIS(vector<int>& nums) {
int temp = 1, ans = 1;
for (int i = 1; i < nums.size(); i++) {
temp = (nums[i-1] < nums[i]) ? temp + 1 : 1;
ans = max(ans, temp);
}
return nums.size() == 0 ? 0 : ans;
}
};
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