从一列数中筛除尽可能少的数使得从左往右看,这些数是从小到大再从大到小的

package cn.xidian.edu.LIS;

import java.util.Arrays;
import java.util.Collections;
import java.util.List;

public class DoubleEndLIS {

	public static void main(String[] args) {
		int array[] = {1,4,5,3,2,1,4,5,6};
		System.out.println(solve(array));
	}

	private static int solve(int[] array) {
		int max = 0;
		for (int i = 1; i < array.length; i++) {
			int[] part1 = Arrays.copyOfRange(array, 0, i);
			int[] part2 = Arrays.copyOfRange(array, i, array.length);
			reverse(part2);
//			System.out.println(Arrays.toString(part2));
			int len1 = solve2(part1);
			int len2 = solve2(part2);
			if (len1+len2 > max) {
				max = len1+len2;
			}
		}
		return array.length-max;
	}
	
	private static void reverse(int[] part2) {
		for (int start = 0, end=part2.length-1; start < end; start++, end--) {
			int t = part2[start];
			part2[start] = part2[end];
			part2[end] = t;
		}
	}

	private static int solve2(int[] nums) {
		//o(n*logn)
		int[] LIS = new int[nums.length+1];
		LIS[1] = nums[0];
		int len = 1;
		for (int i = 1; i < nums.length; i++) {
			int pos = findPos(LIS, 1, len, nums[i]);
			LIS[pos] = nums[i];
			if (len < pos) {
				len = pos;
			}
		}
		return len;
	}
	private static int findPos(int[] lIS, int s, int e, int key) {
		if (key >= lIS[e]) {
			return e+1;
		}
		while (s <= e) {
			int mid = s+(e-s)/2;
			if (lIS[mid] > key) {
				e = mid-1;
			} else if (lIS[mid] < key){
				s = mid+1;
			} else {
				return mid;
			}
		}
		return s;
	}
}

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转载自hcx2013.iteye.com/blog/2218672