LeetCode 34. Search for a Range

LeetCode 34. Search for a Range

Solution1：我的答案
竟然超过了100%，啊哈哈哈哈哈

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

Solution2：
参考网址：http://www.cnblogs.com/grandyang/p/4409379.html
可能有些人会觉得上面的算法不是严格意义上的O(logn)的算法，因为在最坏的情况下会变成O(n)，比如当数组里的数全是目标值的话，从中间向两边找边界就会一直遍历完整个数组，那么我们下面来看一种真正意义上的O(logn)的算法，使用两次二分查找法，第一次找到左边界，第二次调用找到右边界即可，具体代码如下：

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