(二分查找 拓展) leetcode 34. Find First and Last Position of Element in Sorted Array && lintcode 61. Search for a Range

Given an array of integers nums sorted in ascending order, find the starting and ending position of a given target value.

Your algorithm's runtime complexity must be in the order of O(log n).

If the target is not found in the array, return [-1, -1].

Example 1:

Input: nums = [5,7,7,8,8,10], target = 8
Output: [3,4]

Example 2:

Input: nums = [5,7,7,8,8,10], target = 6
Output: [-1,-1]
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这个题是在leetcode和lintcode 上都有的，只不过题目里面给的参数有些是不同的，需要注意的，不过代码是基本相同的。
用遍历数组的方式可能会TLE，所以用二分更好。
emmm，二分查找也可以用递归方式来写的。当用二分查找方式得到一个不为-1数组下标时，向两边扩展，最终得到一个范围。
C++代码：

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        int ans = search(nums,0,nums.size()-1,target);
        if(ans == -1) return {-1,-1};  //归结根底，vector就是一个数组，可以用{}。
        int left = ans,right = ans;
        while(left > 0 && nums[left-1] == nums[ans]) left--;
        while(right < nums.size() - 1 && nums[right+1] == nums[ans]) right++;
        return {left,right};
    }
    int search(vector<int>& nums,int left,int right,int target){
        if(left > right) return -1;
        int mid = left + (right - left)/2;
        if(nums[mid] == target) return mid;
        if(nums[mid] > target) return search(nums,0,mid-1,target);
        else return search(nums,mid+1,right,target);
    }
};

还有一个解法：

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        int ans = search(nums,0,nums.size()-1,target);
        if(ans == -1) return {-1,-1};
        int left = ans,right = ans;
        while(left > 0 && nums[left-1] == nums[ans]) left--;
        while(right < nums.size() - 1 && nums[right+1] == nums[ans]) right++;
        return {left,right};
    }
    int search(vector<int>& nums,int left,int right,int target){
        // if(left > right) return -1;
        // int mid = left + (right - left)/2;
        // if(nums[mid] == target) return mid;
        // if(nums[mid] > target) return search(nums,0,mid-1,target);
        // else return search(nums,mid+1,right,target);
        if(nums.size() == 0) return -1;
        while(left + 1 < right){
            int mid = left + (right - left)/2;
            if(nums[mid] > target) right = mid;
            else if(nums[mid] < target) left = mid;
            else return mid;
        }
        if(nums[left] == target) return left;
        if(nums[right] == target) return right;
        return -1;
    }
};

emmm，也可以看官方题解：https://leetcode.com/articles/find-first-and-last-position-element-sorted-array/