ts实现
function findLengthOfLCIS(nums: number[]): number {  if (nums.length === 1) return 1  let head: number = 0  let next: number = 1  let result: number = 1  while (next !== nums.length) {    nums[next - 1] < nums[next]      ? (result = result <= next - head + 1 ? next - head + 1 : result)      : (head = next)    next++  }  return result}
之前java代码实现
package com.wztlink1013.problems.leetcode.editor.cn;// P674.最长连续递增序列// P674.longest-continuous-increasing-subsequence//给定一个未经排序的整数数组,找到最长且 连续递增的子序列,并返回该序列的长度。 //// 连续递增的子序列 可以由两个下标 l 和 r(l < r)确定,如果对于每个 l <= i < r,都有 nums[i] < nums[i + 1] ,那//么子序列 [nums[l], nums[l + 1], ..., nums[r - 1], nums[r]] 就是连续递增子序列。 //// //// 示例 1: //// //输入:nums = [1,3,5,4,7]//输出:3//解释:最长连续递增序列是 [1,3,5], 长度为3。//尽管 [1,3,5,7] 也是升序的子序列, 但它不是连续的,因为 5 和 7 在原数组里被 4 隔开。 // //// 示例 2: //// //输入:nums = [2,2,2,2,2]//输出:1//解释:最长连续递增序列是 [2], 长度为1。// //// //// 提示: //// // 0 <= nums.length <= 104 // -109 <= nums[i] <= 109 // // Related Topics 数组 // 👍 147 👎 0public class P674LongestContinuousIncreasingSubsequence{    public static void main(String[] args) {        Solution solution = new P674LongestContinuousIncreasingSubsequence().new Solution();        int [] nums = {1,3,5,7};        int result = solution.findLengthOfLCIS(nums);        System.out.println(result);    }//leetcode submit region begin(Prohibit modification and deletion)class Solution {    public int findLengthOfLCIS(int[] nums) {        if (nums.length == 0) { return 0; }        int result = 1;        int count = 1;        for (int i=0; i<nums.length-1; i++) {            if (nums[i] < nums[i+1] ) {                count++;                if (result < count) {result = count;}            } else {                if (result < count) {result = count;}                count = 1;            }        }        return result;    }}//leetcode submit region end(Prohibit modification and deletion)}