19.2.2 [LeetCode 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]

题意

找已排序数组中某个值的范围

题解

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

两次二分