Given an unsorted array of integers, find the length of longest increasing subsequence.Example:
Input: [10,9,2,5,3,7,101,18]
Output: 4
Explanation: The longest increasing subsequence is [2,3,7,101], therefore the length is 4.
Note: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?
难度：medium
题目：给定一无序的整数数组，找出其递增的最大子序列。
思路：动态规划，用一个长度与输入数组相同长度的数组记录从 0 到当前元素的最大递增元素个数。dp(i) = Math.max(dp[j]…) + 1, (nums[i] > nums[j])dp(i) = 1 (nums[i] <= nums[j], 0 <= j <= i – 1)
class Solution {
public int lengthOfLIS(int[] nums) {
int n = nums.length, maxIncLength = 0;
int[] incLength = new int[n];
for (int i = 0; i < n; i++) {
int curMaxIncLength = 0;
for (int j = 0; j <= i; j++) {
if (nums[j] < nums[i]) {
curMaxIncLength = Math.max(curMaxIncLength, incLength[j]);
}
}
incLength[i] = curMaxIncLength + 1;
maxIncLength = Math.max(maxIncLength, incLength[i]);
}
return maxIncLength;
}
}