二分查找LintcodeNo14

14First Position of Target

二分查找的基础题

STL lower_bound实现

class Solution {
public:
    /**
     * @param nums: The integer array.
     * @param target: Target to find.
     * @return: The first position of target. Position starts from 0.
     */
    int binarySearch(vector<int> &nums, int target) {
        // write your code here
        
        auto res_it=lower_bound(nums.begin(),nums.end(),target);
        auto it=nums.begin();
        int i=0;
        if(*res_it != target)return -1;
        return &(*res_it)-&(nums[0]);
        
    }
};

迭代器和下标可使用 取地址符 进行转换 (原理是vector变量在内存上连续分布)

传统算法实现

class Solution {
public:
    /**
     * @param nums: The integer array.
     * @param target: Target to find.
     * @return: The first position of target. Position starts from 0.
     */
    int binarySearch(vector<int> &nums, int target) {
        // write your code here
        
        int left = -1;
        int right = nums.size();
        
        while(right - left > 1)
        {
            int mid = (left+right)/2;
            if(target<=nums[mid])
            {
                right = mid;
            }else{
                left = mid;
            }
            
        }
        if(nums[right] != target)return -1;
        return right;

    }
};

需要注意的是

• 区间为左开 (其原因是 整型变量 (2+3)/2=3 )
• 当搜索到=号的时候,拉右边到中间(right = mid)