题目

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思路

• 二分查找法
• 在找到目标元素时，定义i=j=mid，i–寻找第一次出现的位置，j++寻找最后一次出现的位置

代码

#include<iostream>
#include<vector>
#include<cmath>
#include<algorithm>
using namespace std;
vector<int> searchRange(vector<int>& nums, int target) {
    
    
    vector<int> ret(2);
    ret[0] = ret[1] = -1;
    int size = nums.size();
    int left = 0, right = size - 1;
    while(left <= right) {
    
    
        int mid = (left + right) >> 1;
        if(nums[mid] > target) right = mid - 1;
        else if(nums[mid] < target) left = mid + 1;
        else {
    
    
            int i = mid, j = mid;
            while(i >= 0 && nums[i] == target) {
    
    
                ret[0] = i;
                i--;
            }
            while(j < size && nums[j] == target) {
    
    
                ret[1] = j;
                j++;
            }
            break;
        }
    }
    return ret;
}
//=====官网题解的思路=====================
int pos(vector<int>& nums, int target, bool flag){
    
    
    int size = nums.size();
    int left = 0, right = size - 1;
    while(left <= right) {
    
    
        int mid = (left + right) >> 1;
        if((flag && nums[mid] >= target) || (nums[mid] > target)) 
            right = mid - 1;
        else left = mid + 1;
    }
    return left;
}
vector<int> searchRange1(vector<int>& nums, int target) {
    
    
    vector<int> ret(2);
    ret[0] = ret[1] = -1;
    int pos1 = pos(nums, target, true);
    int pos2 = pos(nums, target, false) - 1;
    if(pos1 <= pos2 && pos1 >= 0 && pos2 < nums.size()){
    
    
        ret[0] = pos1;
        ret[1] = pos2;
    }
    return ret;
}
//========================================
int main(){
    
    
    vector<int> nums;
    nums.emplace_back(1);
    // nums.emplace_back(5);
    // nums.emplace_back(7);
    // nums.emplace_back(7);
    // nums.emplace_back(8);
    // nums.emplace_back(8);
    // nums.emplace_back(10);
    int target = 1;
    vector<int> ret = searchRange1(nums, target);
    cout<<ret[0]<<" "<<ret[1]<<endl;
    return 0;
}