//Given an unsorted array of integers, find the length of longest increasing subsequence.
//For example,
//Given [10, 9, 2, 5, 3, 7, 101, 18],
//The longest increasing subsequence is [2, 3, 7, 101], therefore the length is 4. Note that there may be more than one LIS combination, it is only necessary for you to return the length.
//Your algorithm should run in O(n2) complexity.
//Follow up: Could you improve it to O(n log n) time complexity?
class LongestIncreasingSubsequence {
public int lengthOfLIS(int[] nums) {
if(nums == null || nums.length < 1) {
return 0;
}
int[] dp = new int[nums.length];
dp[0] = 1;
int max = 1;
for(int i = 1; i < dp.length; i++) {
int currentMax = 0;
for(int j = 0; j < i; j++) {
if(nums[i] > nums[j]) {
currentMax = Math.max(currentMax, dp[j]);
}
}
dp[i] = 1 + currentMax;
max = Math.max(max, dp[i]);
}
return max;
}
}
//For example,
//Given [10, 9, 2, 5, 3, 7, 101, 18],
//The longest increasing subsequence is [2, 3, 7, 101], therefore the length is 4. Note that there may be more than one LIS combination, it is only necessary for you to return the length.
//Your algorithm should run in O(n2) complexity.
//Follow up: Could you improve it to O(n log n) time complexity?
class LongestIncreasingSubsequence {
public int lengthOfLIS(int[] nums) {
if(nums == null || nums.length < 1) {
return 0;
}
int[] dp = new int[nums.length];
dp[0] = 1;
int max = 1;
for(int i = 1; i < dp.length; i++) {
int currentMax = 0;
for(int j = 0; j < i; j++) {
if(nums[i] > nums[j]) {
currentMax = Math.max(currentMax, dp[j]);
}
}
dp[i] = 1 + currentMax;
max = Math.max(max, dp[i]);
}
return max;
}
}
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