一，在一个数组里搜索某个元素，最容易被想到的是顺序搜索，从头开始搜，直到搜到为止。
最坏情况运行时间和平均运行时间都和n成线性关系
二，二分法检索的递归算法：
要求数组已经进行了排序，一般按照升序
基本思想：假设数据是按升序排序的，对于给定值x，从序列的中间位置开始比较，如果当前位置值等于x，则查找成功；若x小于当前位置值，则在数列的前半段中查找；若x大于当前位置值则在数列的后半段中继续查找，直到找到为止。<http://zengzhaoshuai.iteye.com/blog/1130376>
public static int binarySearch(Object[ ] a, Object key)
{
// throw new UnsupportedOperationException();
return binarySearch(a,0,a.length-1,key);
} // method binarySearch
private static int binarySearch(Object[] a, int first,int last,Object key)
{
if(first<=last){
int mid=(first+last)>>1;
Object midval=a[mid];
int comp=((Comparable)midval).compareTo(key);
if(comp>0)
return binarySearch(a,first,mid-1,key);
else if(comp==0)
return mid;
else
return binarySearch(a,mid+1,last,key);
}
return -first-1;
} // method binarySearch
成功和不成功搜索的最坏情况运行时间和平均运行时间均和n成对数关系
三，二分法检索的迭代算法
intfirst = 0;
intlast = a.length - 1;
递归算法里的if(first<=last)改成while(first<=last)
//针对int类型数组的二分法查找，key为要查找数的下标
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.
}
四，leetcode上的一系列二分查找的题目
http://oj.leetcode.com/problems/sqrtx/
http://oj.leetcode.com/problems/search-in-rotated-sorted-array/
http://oj.leetcode.com/problems/search-in-rotated-sorted-array-ii/
https://leetcode.com/problems/median-of-two-sorted-arrays/
第一题：
Implement int sqrt(int x).
Compute and return the square root of x.
解法一：二分法class Solution {
public int sqrt(int x) {
if(x == 0) return 0;
if(x==1) return 1;
int start=1, end=x,mid=0;
while(start+1<end){
mid = (start+end)>>>1;
if(mid > x/mid)
end = mid;
else if(mid < x/mid)
start = mid;
else
return mid;
}
return start;
}
};
解法二：牛顿迭代法
转载自http://blog.csdn.net/doc_sgl/article/details/12404971
为了方便理解，就先以本题为例：
计算x2 = n的解，令f(x)=x2-n，相当于求解f(x)=0的解，如左图所示。
首先取x0，如果x0不是解，做一个经过(x0,f(x0))这个点的切线，与x轴的交点为x1。
同样的道理，如果x1不是解，做一个经过(x1,f(x1))这个点的切线，与x轴的交点为x2。
以此类推。
以这样的方式得到的xi会无限趋近于f(x)=0的解。
判断xi是否是f(x)=0的解有两种方法：
一是直接计算f(xi)的值判断是否为0，二是判断前后两个解xi和xi-1是否无限接近。
经过(xi, f(xi))这个点的切线方程为f(x) = f(xi) + f’(xi)(x - xi)，其中f'(x)为f(x)的导数，本题中为2x。令切线方程等于0，即可求出xi+1=xi - f(xi) / f'(xi)。
继续化简，xi+1=xi - (xi2 - n) / (2xi) = xi - xi / 2 + n / (2xi) = xi / 2 + n / 2xi = (xi + n/xi) / 2。
有了迭代公式xi+1= (xi + n/xi) / 2，程序就好写了。关于牛顿迭代法，可以参考wikipedia以及百度百科。
public class Solution {
public int sqrt(int x) {
if (x ==0) return 0;
double pre;
double cur=1;
do
{
pre = cur;
cur = x / (2 * pre) + pre / 2.0;
} while (Math.abs(cur - pre) > 0.00001);
return (int)cur;
}
}
第二题：
Search in Rotated Sorted Array
Suppose a sorted array is rotated at some pivot unknown to you beforehand.
(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).
You are given a target value to search. If found in the array return its index, otherwise return -1.
You may assume no duplicate exists in the array.
public class Solution {
public int search(int[] A, int target) {
// Note: The Solution object is instantiated only once and is reused by each test case.
int start = 0;
int end = A.length-1;
while(start<=end){
int mid =(start+end)>>1;
if(target==A[mid])
return mid;
if(target == A[start])
return start;
if(target == A[end])
return end;
if(A[mid]>A[start]){
if(target>A[start] && target<A[mid])
end = mid -1;
else
start = mid + 1;
}else{
if(target>A[mid] && target<A[end])
start = mid+1;
else
end = mid-1;
}
}
return -1;
}
}
第三题：
Search in Rotated Sorted Array II
Follow up for "Search in Rotated Sorted Array":
What if duplicates are allowed?
Would this affect the run-time complexity? How and why?
Write a function to determine if a given target is in the array.
public class Solution {
public boolean search(int[] A, int target) {
if(A==null) throw new IllegalArgumentException();
int start=0,end=A.length-1,mid;
while(start<=end){
mid =(end+start)>>1;
if(target==A[mid]||target == A[start]||target == A[end])
return true;
if(A[mid]>A[start]){//mid在左数组，分界线在mid右边
if(target>A[start] && target<A[mid])
end = mid -1;
else
start = mid + 1;
}else if(A[mid]<A[start]){ //mid在右数组，分界线在mid左边
if(target>A[mid] && target<A[end])
start = mid+1;
else
end = mid-1;
}
else{
start++;
}
}
return false;
}
}
第四题
Median of Two Sorted Arrays
There are two sorted arrays nums1 and nums2 of size m and n respectively.
Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).
Example 1:
nums1 = [1, 3]
nums2 = [2]
The median is 2.0
Example 2:
nums1 = [1, 2]
nums2 = [3, 4]
The median is (2 + 3)/2 = 2.5
解答：
https://discuss.leetcode.com/topic/4996/share-my-o-log-min-m-n-solution-with-explanation/15
<pre name="code" class="java">public class Solution {
public double findMedianSortedArrays(int[] nums1, int[] nums2) {
int m=nums1.length,n=nums2.length;
if(m>n){
return findMedianSortedArrays(nums2,nums1);
}
int i=0,j=0,imin=0,imax=m,half=(m+n+1)>>1;
int maxLeft=0,minRight=0;
while(imin<=imax){
i=(imin+imax)>>1;
j=half-i;
if(i>0&&j<n&&nums1[i-1]>nums2[j]){
imax=i-1;
}else if(j>0&&i<m&&nums2[j-1]>nums1[i]){
imin=i+1;
}else{
if(i==0) maxLeft=nums2[j-1];
else if(j==0) maxLeft=nums1[i-1];
else maxLeft=getMax(nums1[i-1],nums2[j-1]);
break;
}
}
//如果m+n为奇数，取中间值
if(((m+n)&1)!=0) return maxLeft;
if(i==m) minRight=nums2[j];
else if(j==n) minRight=nums1[i];
else minRight=getMin(nums1[i],nums2[j]);
return ((double)(maxLeft+minRight))/2;
}
public int getMax(int a,int b){
return a>b?a:b;
}
public int getMin(int a,int b){
return a<b?a:b;
}
}