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leetcode (Longest Continuous Increasing Subsequence)

Title:Longest Continuous Increasing Subsequence    674

Difficulty:Easy

原題leetcode地址:https://leetcode.com/problems/longest-continuous-increasing-subsequence/

 

1.  後面的數大每次加加,否則為1,每次都比較下大小

時間複雜度:O(n),一次一層for迴圈,需要遍歷整個陣列。

空間複雜度:O(1),沒有申請額外的空間。

    /**
     * 後面的數大每次加加,否則為1,每次都比較下大小
     * @param nums
     * @return
     */
    public static int findLengthOfLCIS(int[] nums) {

        if (nums == null || nums.length <= 0) {
            return 0;
        }

        if (nums.length == 1) {
            return 1;
        }

        int count = 1;
        int max = Integer.MIN_VALUE;

        for (int i = 0; i < nums.length - 1; i++) {
            if (nums[i + 1] > nums[i]) {
                count++;
            }
            else {
                count = 1;
            }
            if (count > max) {
                max = count;
            }
        }

        return max;

    }

2.  動態規劃

時間複雜度:O(n),一次一層for迴圈,需要遍歷整個陣列。

空間複雜度:O(n),申請與nums長度一樣的大的陣列長度n。

    /**
     * 動態規劃
     * @param nums
     * @return
     */
    public static int findLengthOfLCIS1(int[] nums) {

        if (nums == null || nums.length <= 0) {
            return 0;
        }

        if (nums.length == 1) {
            return 1;
        }

        int tmp[] = new int[nums.length];
        tmp[0] = 1;

        for (int i = 1; i < nums.length; i++) {
            tmp[i] = 1;
            if (nums[i] > nums[i - 1]) {
                tmp[i] = tmp[i - 1] + 1;
            }
        }

        int max = Integer.MIN_VALUE;

        for (int i = 0; i < tmp.length; i++) {
            if (tmp[i] > max) {
                max = tmp[i];
            }
        }

        return max;

    }