算法分析与设计课程——LeetCode刷题之Search for a Range

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题目:

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].


答案:

vector<int> searchRange(vector<int>& nums, int target) {
    int idx1 = lower_bound(nums, target);
    int idx2 = lower_bound(nums, target+1)-1;
    if (idx1 < nums.size() && nums[idx1] == target)
        return {idx1, idx2};
    else
        return {-1, -1};
}

int lower_bound(vector<int>& nums, int target) {
    int l = 0, r = nums.size()-1;
    while (l <= r) {
        int mid = (r-l)/2+l;
        if (nums[mid] < target)
            l = mid+1;
        else
            r = mid-1;
    }
    return l;
}


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转载自blog.csdn.net/m0_38041038/article/details/79039087