Find First and Last Position of Element in Sorted Array

目录

• 题目链接
• 注意点
• 解法
• 小结

题目链接

Find First and Last Position of Element in Sorted Array - LeetCode

注意点

• nums可能为空
• 时间复杂度为O(logn)

解法

解法一：最普通的二分搜索，先找到一个target，然后向两边拓展。

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

解法二：解法一在最坏情况下时间复杂度是O(n),比如整个nums都是target。因此我们用两次二分搜索，第一次找到第一个等于target的数字，第二次找到最后一个等于target的数字。时间复杂度O(logn)

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

小结

• 二分搜索变种题