lintcode:First Position of Target_lintcoude中first position of target-优快云博客

本文介绍两种二分查找方法实现：搜索左闭右闭区间及搜索左闭右开区间，通过这两种方式可在已排序数组中找到目标值的第一个出现位置，若不存在则返回-1。文章提供C++代码实例。

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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.

Example
If the array is [1, 2, 3, 3, 4, 5, 10], for given target 3, return 2.

1.方法1：搜索左闭右闭区间

注意求中间值尽量避免使用（l+r）/2，因为l+r可能会溢出；
所以尽量用l+(r-l)/2这样就中间值；

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

        return -1;
    }
};

2.方法2：搜索左闭右开空间

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

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

        if(array[l]==target){
            return l;
        }

        return -1;
    }
};