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;}
};

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