[Leetcode] Increasing Triplet Subsequence 递增的三元子序列

Given an unsorted array return whether an increasing subsequence of
length 3 exists or not in the array. More specifically, if there exists i , j , k such that arr[i] < arr[j] < arr[k] given 0 ≤ i < j < k ≤ n-1 return true else return false . Your function should run in O(n) time complexity and O(1) space complexity.
Examples:
Given [1, 2, 3, 4, 5] , return true .
Given [5, 4, 3, 2, 1] , return false .

实现代码
IncreasingTripletSubsequence.java

package array;

import java.util.Arrays;

import util.Print;

public class IncreasingTripletSubsequence {
    /**
     * 描述
        Given an unsorted array return whether an increasing subsequence of length 3 exists or not in the array.
        More specifically, if there exists i , j , k such that arr[i] < arr[j] < arr[k] given 0 ≤ i < j
        < k ≤ n-1 return true else return false .
        Your function should run in O(n) time complexity and O(1) space complexity.
        Examples:
        Given [1, 2, 3, 4, 5] , return true .
        Given [5, 4, 3, 2, 1] , return false .
     * 分析
         对一个无序数组，判读递增3数列是否存在
     * 直接解法
         扫描数组，遍历三遍
     * 复杂度
         时间(n^3),空间 (1)
     * @param nums
     * @return
     */
    public boolean Solution1(int[] nums){
        
        int min=nums[0];
        
        for(int i=0;i<nums.length;i++)
            for(int j=i+1; j<nums.length;j++){
                if(nums[i] >= nums[j])
                    continue;
                for(int k=j+1;k<nums.length;k++){
                    if(nums[k]>nums[j]){
                        Print.Int(nums[i]);
                        Print.Int(nums[j]);
                        Print.Int(nums[k]);
                        return true;    
                    }
                }
            }
        
        return false;        
    }
    
    /**
     * 夹逼解法
         对每个i用j,k夹逼出结果
     * 复杂度
         时间O(n^2),,空间(1)
     * @param nums
     * @return
     */
    public boolean Solution2(int[] nums){
                
        for(int i=0;i<nums.length;i++){
            int j = i+1;
            int k = nums.length-1;
            while(j<k){
                while(nums[i]>=nums[j] && j<k)    j++;
                while(nums[j]>=nums[k] && j<k ) k--;
                if(j<k) {
                    Print.Int(nums[i]);
                    Print.Int(nums[j]);
                    Print.Int(nums[k]);
                    return true;
                }
                //else break;
            }
        }
            
        
        return false;        
    }
        
/**
 * 时间优化解法
 * 分析
    扫描一遍数组，用变量 x1 保存当前最小的值，变量 x2 保存当前第二小的值。若是右面能碰到一个数大于 x2 ，说明必然存在一个递增的三元组。
 * 复杂度
     空间(1),时间(n)
 */

public boolean Solution3(int[] nums){
    int x1=Integer.MAX_VALUE, x2=Integer.MAX_VALUE;
    
    for(int i=0;i<nums.length;i++){
        if(nums[i]<x1) x1=nums[i];
        else if (nums[i]<x2) x2=nums[i];
        else return true;
        }
    return false;
     
    }

}

测试代码
IncreasingTripletSubsequenceTest.java

package array;

import org.junit.Assert;
import org.junit.Before;
import org.junit.Test;

public class IncreasingTripletSubsequenceTest {

    private IncreasingTripletSubsequence s;
    
    @Before
        public void setUp() {
         s = new IncreasingTripletSubsequence();
        }
     
    @Test
    public void testSolution1() {
        
        int[] nums = {3,4,1,7,5,2};
        boolean expect = true;
        
        boolean result = s.Solution1(nums);
        System.out.println(result);
        Assert.assertEquals(expect, result);

    }

    @Test
    public void testSolution2() {
        
        int[] nums ={9,1,6,8,7};
        boolean expect = true;
        
        boolean result = s.Solution2(nums);
        System.out.println(result);
        Assert.assertEquals(expect, result);

    }
    @Test
    public void testSolution3() {
        
        int[] nums ={5,4,3,2,1};
        boolean expect = false;
        
        boolean result = s.Solution3(nums);
        System.out.println(result);
        Assert.assertEquals(expect, result);

    }
    
}

结果

3 4 7 true
1 6 7 true
false