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Data Structures and Algorithm Interview Questions from the internet. Help programmers to get hired by dream companies like Google or Search,这道题使用的二分搜索法是博主归纳总结帖LeetCode Binary Search Summary 二分搜索法小结 我们初始化left = 0, right = 1,作为二分查找的边界。然后每次取left和right的中间值,再分别计算每行中有多少个值小于等于mid。最后将所有行中小于等于mid的分数的个数加起来,记为count。如果发现count小于K,那么说明mid的取值过小,所以修改左边界;否则修改右边界。这样当count的个数刚好为K的时候,我们取小于等于mid的所有分数中的最大值,即为第K大的分数。二分查找法的空间复杂度为O(1),时间复杂度为O(log(r)n^2),其中r是left和right之间的差值。可以发现这个算法的时间复杂度与K无关。由于K一般情况下下是O(n^2)量级的,所以二分查找的时间复杂度还是要好于基于优先队列的方法。     vector kthSmallestPrimeFraction(vector& A, int K) {         double left = 0, right = 1;         int p = 0, q = 1, cnt = 0, n = A.size();         while (true) {                 }             }             if (cnt == K) return {p, q};             else if (cnt < K) left = mid;             else right = mid;         }       if (n == j) break;       total += (n - j);     if (pq.isEmpty()) {         throw new RuntimeException(invalid input.); This solution probably doesn't have the best runtime but it's really simple and easy to understand. So now the problem becomes find the kth smallest element of (n-1) sorted list                 return s1 - s2;             }                 pq.add(pop);             }


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