摘要：Note: 这道题很有意思，用的是二分法。想法是不停的试错。这个博客主把这个问题说的很清楚 https://xuezhashuati.blogspot.com/2017/03/lintcode-437-copy-books.html 其实二分法的方法可以变化的点很有限，最多变化的就是检查条件。一般难 阅读全文

摘要：Given a set of distinct positive integers, find the largest subset such that every pair (Si, Sj) of elements in this subset satisfies: Si % Sj = 0 or  阅读全文

摘要：Given a sequence of integers, find the longest increasing subsequence (LIS). You code should return the length of the LIS. Given a sequence of integer 阅读全文

摘要：方法一： 动态规划 https://segmentfault.com/a/1190000003768736 https://siddontang.gitbooks.io/leetcode-solution/content/dynamic_programming/perfect_squares.htm 阅读全文

摘要：Important Note:1) Pay attention to the number that is less than 1.0, because x^0.5 will larger than x. So we need to always set the right side to be 1 阅读全文

摘要：Important Reference: http://stackoverflow.com/questions/13093602/finding-subarray-with-maximum-sum-number-of-elements http://www.jiuzhang.com/qa/2942/ 阅读全文

摘要：The reference link: http://www.code123.cc/docs/leetcode-notes/binary_search/search_a_2d_matrix_ii.html 题解 - 自右上而左下 阅读全文

摘要：This question is very hard, just remember it:Explaination: 思路清晰，就是二倍法。直接用除数去一个一个加，直到被除数被超过的话，会超时。解决办法每次将被除数增加1倍，同时将count也增加一倍，如果超过了被除数，那么用被除数减去当前和再继续本 阅读全文

摘要：Important Node: 1) This problem has two parts: a. find the closest number's to the target and return its index. This is standard binary search problem 阅读全文

摘要：The worst situation O(N). Actually we can either just loop through, or we can compare num[mid] with the num[end], if they are the same, that means it' 阅读全文

摘要：O(logN) This question turns to find the first and last element of the target in a sorted array. Just be careful with the two result coming out of the 阅读全文

摘要：O(N) if element can repeat, the worst case, you cannot throw away any section. eg. [1, 1, 1, 1, 0, 1, 1] target = 0, you cannot throw any section. We 阅读全文

摘要：O(logN) Important Point: Once the target is in one section, use the point in that section as benchmark In this problem, if the target >= startVal, use 阅读全文

摘要：O(logN) 阅读全文

摘要：O(logN) Check four different situations: 1) The peek 2) increasing section 3) decreasing section 4) minimum 阅读全文

摘要：O(logN) There are two section: 1) increase array which the first element is larger than end one. 2) the minimum to end Comparing with the endVal, 1) i 阅读全文

摘要：O(logN) 阅读全文

摘要：O(logN) Important Point: if there "mid" exists, it has at laest three elements in "nums", because otherwise it will not come inside the loop. So later 阅读全文

摘要：class Solution { /** * @param x: An integer * @return: The sqrt of x */ public int sqrt(int x) { // write your code here if (x < 0) { return -1; ... 阅读全文

摘要：If it is two eggs: http://datagenetics.com/blog/july22012/index.html Imagine we drop our first egg from floor n, if it breaks, we can step through the 阅读全文