674. 最长连续递增序列

• dp 时间复杂度O(N), 空间复杂度O(N)

    /**
     *  nums [1, 3, 5, 4, 7]
     *                 dp
     *  [1] ->         1
     *  [1,3] ->       2
     *  [1,3,5] ->     3
     *  [1,3,5,4] ->   1
     *  [1,3,5,4,7] -> 2
     *
     *  maxLen -> 3
     */
    public int findLengthOfLCIS(int[] nums) {
        int[] length= new int[nums.length];
        Arrays.fill(length, 1);
        for (int i = 1; i < nums.length; i++) {
            if(nums[i] > nums[i-1]) {
                length[i] = length[i-1] + 1;
            }
        }
        int maxLen = 0;
        for (int len : length) {
            maxLen = Math.max(maxLen, len);
        }
        return maxLen;
    }

• dp 优化 时间复杂度O(N), 空间复杂度O(1)

    public int findLengthOfLCIS(int[] nums) { //类似滑动窗口
        if (nums.length<=0) return 0;
        int before = 1;
        int maxLen = 1;
        for (int i = 1; i < nums.length; i++) {
            if(nums[i] > nums[i-1]) {
                before = before + 1;
                maxLen = Math.max(maxLen, before);
            } else {
                before = 1;
            }
        }
        return maxLen;
    }