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Divide two integers without using multiplication, division and mod operator. If it is overflow, return MAX_INT. 法一：可以直观的利用除数的倍数，但如果这个倍数很大（题目中要求不能用乘法）则会造成LTE， 此法不可行！！！ class Solution { public

Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand. (i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2). You are given a target value to search. If foun

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

Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order. You may assume no duplicates in the array.

Implement pow(x, n). 需要注意的两点 1， 当n为 INT_MIN时： int n = INT_MIN; int a = -n; // a == -22147483648 int b = abs(n);// b == -22147483648 2，n 右移了多少次， x 翻多少翻，而不是ans翻多少翻，因为还有ans = ans * x； AC 代码：