Leetcode-674 Longest Continuous Increasing

Longest Continuous Increasing

题目描述

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

Example

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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.
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

解法1

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,我们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。我们用一个变量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:

cpp

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

Swift

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class Solution {
func findLengthOfLCIS(_ nums: [Int]) -> Int {
var res = 0, cnt = 0, pre = Int.max
for num in nums {
if pre < num { cnt += 1 }
else { cnt = 1 }
res = max(res, cnt)
pre = num
}
return res
}
}

解法2

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:

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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;
}
}

Swift

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class Solution {
func findLengthOfLCIS(_ nums: [Int]) -> Int {
var res = 0, cnt = 0, n = nums.count
for i in 0..<n {
if i == 0 || nums[i - 1] < nums[i] {
cnt += 1
res = max(res, cnt)
} else {
cnt = 1
}
}
return res
}
}

类似题目

Number of Longest Increasing Subsequence

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