算法修炼Day52|● 300.最长递增子序列 ● 674. 最长连续递增序列 ● 718. 最长重复子数组

 1class Solution {
 2    public int lengthOfLIS(int[] nums) {
 3        int[] dp = new int[nums.length];
 4        Arrays.fill(dp, 1);
 5        for (int i = 1; i < nums.length; i++) {
 6            for (int j = 0; j < i; j++) {
 7                if (nums[j] < nums[i]) {
 8                    dp[i] = Math.max(dp[j] + 1, dp[i]);
 9                }
10            }
11        }
12        int res = 0;
13        for (int i = 0; i < nums.length; i++) {
14            res = Math.max(res, dp[i]);
15        }
16        return res;
17    }
18}

 1class Solution {
 2    public int findLengthOfLCIS(int[] nums) {
 3        int[] dp = new int[nums.length];
 4        Arrays.fill(dp, 1);
 5
 6        for (int i = 0; i < nums.length - 1; i++) {        
 7
 8            if (nums[i + 1] > nums[i]) {
 9                dp[i + 1] = dp[i] + 1;
10            }
11        }
12        int res = 0;
13        for (int i = 0; i < dp.length; i++) {
14            res = Math.max(dp[i], res);
15        }
16        return res;
17    }
18}

 1// 暴力解法
 2class Solution {
 3    public int findLength(int[] nums1, int[] nums2) {
 4        int maxLength = 0;
 5        for (int i = 0; i < nums1.length; i++) {
 6            for (int j = 0; j < nums2.length; j++) {
 7
 8                int length = 0;
 9                int p1 = i;
10                int p2 = j;
11
12                while (p1 < nums1.length && p2 < nums2.length && nums1[p1] == nums2[p2]) {
13                    length++;
14                    p1++;
15                    p2++;
16                }
17                maxLength = Math.max(maxLength, length);
18            }
19        }
20        return maxLength;
21    }
22}
23
24// 动规
25class Solution {
26    public int findLength(int[] nums1, int[] nums2) {
27        int res = 0;
28        int[][] dp = new int[nums1.length + 1][nums2.length + 1];
29
30        for (int i = 1; i < nums1.length + 1; i++) {
31            for (int j = 1; j < nums2.length + 1; j++) {
32                if (nums1[i - 1] == nums2[j - 1]) {
33                    dp[i][j] = dp[i - 1][j - 1] + 1;
34                    res = Math.max(res, dp[i][j]); 
35                }
36            }
37        }
38        return res;
39    }
40}