674. Longest Continuous Increasing Subsequence
674. Longest Continuous Increasing Subsequence
用户1147447
发布于 2019-05-26 00:19:20
发布于 2019-05-26 00:19:20
LWC 49：674. Longest Continuous Increasing Subsequence
Problem:
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.
思路： 用了DP记录当前序列的最大连续长度。
dp[i] 表示当前位置的最大连续长度
更新：
dp[i] = dp[i - 1] + 1 if (前一元素小于当前元素)代码如下：
public int findLengthOfLCIS(int[] nums) {
if (nums.length == 0) return 0;
int[] dp = new int[nums.length];
Arrays.fill(dp, 1);
int max = 1;
for (int i = 1; i < nums.length; ++i) {
if (nums[i] > nums[i - 1]) {
dp[i] = dp[i - 1] + 1;
}
max = Math.max(max, dp[i]);
}
return max;
}此题不一定需要DP，直接用单个变量控制即可。
代码如下：
public int findLengthOfLCIS(int[] nums) {
int n = nums.length;
if (n == 0) return 0;
int max = 1;
for (int i = 1, k = 1; i < n; ++i) {
if (nums[i] > nums[i - 1]) {
k ++;
max = Math.max(max, k);
}
else {
k = 1;
}
}
return max;
}本文参与 腾讯云自媒体分享计划，分享自作者个人站点/博客。
原始发表：2017年09月10日，如有侵权请联系 cloudcommunity@tencent.com 删除
评论
登录后参与评论
推荐阅读
目录