题目描述

(Medium)

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

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

You must write an algorithm with O(log n) runtime complexity.

Example 1:

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

Example 2:

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Input: nums = [5,7,7,8,8,10], target = 6
Output: [-1,-1]

Example 3:

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Input: nums = [], target = 0
Output: [-1,-1]

思路

1 二分查找

代码

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

        return ans;
    }

    int findLast(vector<int>& nums, int target){
        int left = 0, right = nums.size() -1, ans = -1;
        while(left <= right){
            int mid = left + (right - left) / 2;
            if(nums[mid] == target){
                ans = mid;
                left = mid + 1; //keep searching the right;
            }else if (nums[mid] < target){
                left = mid + 1;
            }else{
                right = mid -1;
            }
        }

        return ans;
    }


    vector<int> searchRange(vector<int>& nums, int target) {
        int n = nums.size();
        //vector<int> re;

        if(n  == 0){
            vector<int> v = {-1, -1};
            return v;
        }

        int left = findFirst(nums, target);
        int right = findLast(nums, target);


        return {left, right};
    }
};