简单二分法-First Position of Target

Problem 7：

For a given sorted array (ascending order) and a 

target

number, find the first index
of this number in 

O(log n)

 time complexity.

If the target number does not exist in the array, return 

-1

.

数组设置start,end,mid变量进行index循环比较，start=end时也就是指的是同一个元素时跳出循环，此时判断array[start]是否与target一致：

class Solution {
public:
/**
* @param nums: The integer array.
* @param target: Target number to find.
* @return: The first position of target. Position starts from 0.
*/
int binarySearch(vector<int> &array, int target) {
// write your code here
//Using binary search
if(array.empty()||array.size() == 0 )
return -1;

int start = 0, end = array.size()-1;
while(start < end){
int mid = start + (end - start)/2;
if(array[mid] == target)
end = mid;
else if  (array[mid] < target)
start = mid +1;
else
end = mid -1;
}
if(array[start] == target)
return start;
return -1;
}
};

  若当start = end -1 时跳出循环，此时start与end 相邻，需要判断start和end两个位置对应的值是否与target一致：

class Solution {
public:
/**
* @param nums: The integer array.
* @param target: Target number to find.
* @return: The first position of target. Position starts from 0.
*/
int binarySearch(vector<int> &array, int target) {
// write your code here
//Using binary search
if(array.empty()||array.size() == 0 )
return -1;

int start = 0, end = array.size() -1;
while(start +1 < end){
int mid = start + (end - start)/2;
if(array[mid] == target)
end = mid;
else if  (array[mid] < target)
start = mid;
else
end = mid;
}
if(array[start] == target)
return start;
if(array[end] == target)
return end;
return -1;
}
};

上述两个方法是迭代实现，如果要递归实现：

class Solution {
/**
* @param nums: The integer array.
* @param target: Target to find.
* @return: The first position of target. Position starts f
4000
rom 0.
*/
public int binarySearch(int[] nums, int target) {
//write your code here
if(nums.length==1&&target==nums[0])   return 0;
int len = nums.length;
int index = binaryFind(nums,target,0,len);
return index;
}
public  int binaryFind(int[] nums,int target,int left,int right){
int median = (left+right)/2;
if(nums[left]==target)  return left;
if(left >= right )  return -1;
if(target == nums[median]&&target != nums[median-1])  return median;//获取最前匹配位置
if(target > nums[median])  return binaryFind(nums,target,median+1,right);
if(target <= nums[median])  return binaryFind(nums,target,left,median);
return -1;
}

}

递归算法参照http://www.cnblogs.com/wangnanabuaa/p/4987304.html