二分法查找的效率

结果:排序需要耗费巨大时间。单纯二分查找需要时间很少，其空间复杂度为O(1)，时间复杂度为O(logN),而普通查找的时间复杂度为O(N)，空间复杂度也为O(1)。

测试数据使用python代码生成，

#coding=utf-8
import random;
fp=open("test.txt","w+");
n=50000000;
k=50000000;
for i in range(1,n):
    r=random.randint(1,k);
    fp.write(str(r)+"\n");
fp.close();

测试java代码如下，

public class Demo {
	public static void main(String[] args) {
		String dstFile = args[0];
//		System.out.println(dstFile);
		
		int[] a = readInts("test.txt");
		
		int[] b = readInts("search.txt");
		
		long start = System.currentTimeMillis();
		
//		for(int i=0;i<b.length;i++){
//			if(find(a,b[i])){
//				System.out.println(b[i] + " find!");
//			}
//		} 
		
		long s1=System.currentTimeMillis();
		Arrays.sort(a);  // 排序时间比较多，查找时间是相当短地，经过测试对于5千万的数据，
		long s2=System.currentTimeMillis();
		System.out.println("the sort cost time is " + (s2-s1)/1000.0 + "s");
		
		long ss=System.currentTimeMillis();
		for(int i=0;i<b.length;i++){
			if(binaryQuery(a,b[i]) != -1){
				System.out.println(b[i] + " find!");
			}
		} 
		long ee=System.currentTimeMillis();
		System.out.println("the binary find cost time is " + (ee-ss)/1000.0 + "s");
		
		long end = System.currentTimeMillis();
		System.out.println("the cost time is " + (end-start)/1000.0 + "s");
		
//		List<int[]> list = Arrays.asList(a);
//		System.out.println(list.);
		
//		File f = new File();
//		FileInputStream
	}
	
	private static int[] readInts(String path){
		File f = new File(path);
		List<Integer> list = new ArrayList<Integer>();
		BufferedReader br = null;
		try {
			br = new BufferedReader(new FileReader(f));
			String temp = null;
			while((temp = br.readLine()) != null){
				int k = Integer.valueOf(temp);
//				System.out.println(temp);
				list.add(k);
			}
		} catch (FileNotFoundException e) {
			e.printStackTrace();
		} catch (IOException e) {
			e.printStackTrace();
		} finally {
			if(br != null){
				try {
					br.close();
				} catch (IOException e) {
					e.printStackTrace();
				}
				br = null;
			}
		}
		
//		String[] array = (String[])list.toArray(new String[list.size()]);
//		int[] ret = new int[list.size() - 1];
//		for(int i=0;i<array.length - 1;i++){
//			ret[i] = Integer.valueOf(array[i]);
//		}
		Integer[] tempRet = list.toArray(new Integer[list.size()]);
		int[] ret = new int[list.size()];
		for(int i=0;i<tempRet.length;i++){
			ret[i] = tempRet[i];
		}
		return ret;
	}
	
	private static boolean find(int[] a,int x){
		int k = rawQuery(a, x);
		if(k == -1){
			return false;
		}
		return true;
	}
	
	private static int rawQuery(int[] a,int x){
		for(int i=0;i<a.length;i++){
			if(x==a[i]){
				return i;
			}
		}
		return -1;
	}
	
	private static int binaryQuery(int[] a,int x){
		int l=0,h=a.length-1;
		while(l <= h){
			int mid = (l+h)/2;
			if(x < a[mid]){
				h = mid - 1;
			} else if(x > a[mid]){
				l = mid + 1;
			} else{
				return mid;
			}
		}
		return -1;
	}
}