Leetcode 34 Search for a Range

Given an array of integers 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].

For example,
Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].

解题思路：采用分治的思想对数组进行折半查找，这题重点在怎么找到target在数组的左端点的位置、右端点的位置，比如判断左端点就是找对应target的下标的前一个位置所对应的值，是否小于target；同理右端点也是这样思路，然后再分别判断一下各自边界条件是啥。

#include <stdio.h>

#include <vector>

int left_bound(std::vector<int>& nums, int target){
    int begin = 0;
    int end = nums.size() - 1;
    while(begin <= end){
        int mid = (begin + end) / 2;
        if (target == nums[mid]){
            if (mid == 0 || nums[mid -1] < target){
                return mid;
            }
            end = mid - 1;
        }
        else if (target < nums[mid]){
            end = mid - 1;
        }
        else if (target > nums[mid]){
            begin = mid + 1;
        }
    }
    return -1;
}

int right_bound(std::vector<int>& nums, int target){
    int begin = 0;
    int end = nums.size() - 1;
    while(begin <= end){
        int mid = (begin + end) / 2;
        if (target == nums[mid]){
            if (mid == nums.size() - 1 || nums[mid + 1] > target){
                return mid;
            }
            begin = mid + 1;
        }
        else if (target < nums[mid]){
            end = mid - 1;
        }
        else if (target > nums[mid]){
            begin = mid + 1;
        }
    }
    return -1;
}

class Solution {
public:
    std::vector<int> searchRange(std::vector<int>& nums, int target) {
        std::vector<int> result;
        result.push_back(left_bound(nums, target));
        result.push_back(right_bound(nums, target));
        return result;
    }
};

int main(){
    int test[] = {5, 7, 7, 8, 8, 8, 8, 10};
    std::vector<int> nums;
    Solution solve;
    for (int i = 0; i < 8; i++){
        nums.push_back(test[i]);
    }
    for (int i = 0; i < 12; i++){
        std::vector<int> result = solve.searchRange(nums, i);
        printf("%d : [%d, %d]
",i , result[0], result[1]);
    }
    return 0;
}

通过~