0034. 在排序数组中查找元素的第一个和最后一个位置 [Medium] [Find First and Last Position of Element in Sorted Array].md

34. Find First and Last Position of Element in Sorted Array

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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题目：有序数组的目标数对应的区间上下范围。

思路：二分查找。参考Discuss。

工程代码下载

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        int n = nums.size();
        if(n == 0) return {-1, -1};
        int l = 0, r = n-1;
        while(l < r){
            int mid = l + (r-l)/2;
            if(nums[mid] > target)
                r = mid - 1;
            else if(nums[mid] < target)
                l = mid + 1;
            else
                r = mid;
        }
        if(nums[l] != target)
            return {-1, -1};
        vector<int> res(2);
        res[0] = l;

        r = n-1; // 找右侧边界时l不用赋值为0
        while(l < r){
            int mid = l + (r-l)/2 + 1;  // 注意这里多加了1
            if(nums[mid] > target)
                r = mid - 1;
            else if(nums[mid] < target)
                l = mid + 1;
            else
                l = mid;
        }
        res[1] = r;
        return res;
    }
};

另可参考解法2

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        int n = nums.size();
        if(n == 0) return {-1, -1};
        double left = target - 0.5, right = target + 0.5;
        int l = binarySearch(nums, left);
        int r = binarySearch(nums, right);
        if(l == r) return {-1, -1};
        return {l, r-1};
    }
private:
    int binarySearch(const vector<int>& nums, double target){
        int n = nums.size();
        int l = 0, r = n-1;
        while(l <= r){
            int mid = l + (r - l)/2;
            if(target > nums[mid])
                l = mid + 1;
            else
                r = mid - 1;
        }
        return l;
    }
};