LeetCode 300 Longest Increasing Subsequence

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?

方法一：动态规划。O(n^2)解法（运行时间33
ms），参考：西施豆腐渣

public int lengthOfLIS(int[] nums) {
int max = 0;
int[] dp = new int[nums.length];//d[i]为subset0...i的Longest increasing sub.
for (int i = 0; i < nums.length; i++) {
dp[i] = 1;
for (int j = 0; j < i; j++) {
if (nums[i] > nums[j]) dp[i] = Math.max(dp[i], dp[j] + 1);
}
max = Math.max(max, dp[i]);
}
return max;
}

方法二: O(n * log n)解法（运行时间1ms）：

维护一个单调序列。遍历nums数组，二分查找每一个数在单调序列中的位置，然后替换。

public int lengthOfLIS2(int[] nums) {
int[] dp = new int[nums.length];
int len = 0;
for (int x : nums) {
int i = Arrays.binarySearch(dp, 0, len, x);//二分查找x在单调序列的位置
if (i < 0) i = -(i + 1);
dp[i] = x;
if (i == len) len++;
}
return len;
}

另外，Arrays.binarySearch(dp, 0, len, x)的底层实现为：

// Like public version, but without range checks.
private static int binarySearch0(int[] a, int fromIndex, int toIndex,
int key) {
int low = fromIndex;
int high = toIndex - 1;

while (low <= high) {
int mid = (low + high) >>> 1;
int midVal = a[mid];

if (midVal < key)
low = mid + 1;
else if (midVal > key)
high = mid - 1;
else
return mid; // key found
}
return -(low + 1);  // key not found.
}