简单二分法-First Position of Target

Problem 7：

For a given sorted array (ascending order) and a targetnumber, 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 from 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