leetcode题解-153. Find Minimum in Rotated Sorted Array && 238. Product of Array Except Self

153， 题目：

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).Find the minimum element.You may assume no duplicate exists in the array.

本题和之前的一道题目很相似，都是将排序数组在某处进行翻转，不同的是本题寻找最小值。思路也很简单，两种方法：直接遍历和binary search。代码入下：

    public int findMin(int[] nums) {        for(int i=0; i<nums.length-1; i++){            if(nums[i] > nums[i+1])                return nums[i+1];        }        return nums[0];    }    public  int findMin1(int[] nums){        int left=0, right=nums.length-1, mid;        while(left < right){            mid = (left + right)/2;            if(nums[mid] > nums[right]) {                if(mid < nums.length-1 && nums[mid] > nums[mid+1])                    return nums[mid+1];                left = mid+1;            }            else{                if(mid >0 && nums[mid] < nums[mid-1])                    return nums[mid];                right = mid-1;            }        }        return nums[0];    }    public int findMin2(int[] nums) {        int start = 0, end = nums.length - 1, mid;        while (start  <  end) {            mid = (start + end) / 2;            if (nums[mid]  > nums[end])                start = mid + 1;            else                end = mid;        }        return nums[start];    }

238，题目：

Given an array of n integers where n > 1, nums, return an array output such that output[i] is equal to the product of all the elements of nums except nums[i].Solve it without division and in O(n).For example, given [1,2,3,4], return [24,12,8,6].Follow up:Could you solve it with constant space complexity? (Note: The output array does not count as extra space for the purpose of space complexity analysis.)

本题是求元素乘积（除了该位置元素外的其他所有元素）。很简单，一开始想到的方法比如，求所有元素之积然后除，背题目限制。然后还可以暴力遍历每个元素对其他元素求积，但是会超过时间限制。最后想到可以使用两次遍历的方法，第一次遍历得到该元素之前的所有元素之积，然后第二次进行反向遍历，用于求除去该元素之外所有元素之积。例如:

[1,2,3,4,5,6,7]第一次：[1,1,2,6,24,120,720]第二次：[....6*1*7*6*5, 24*1*7*6, 120*1*7, 720*1]

代码入下：

//暴力法    public  int[] productExpectSelf1(int[] nums){        int[] res = new int[nums.length];        for(int i=0; i<nums.length; i++){            int tmp = 1;            for(int j=0; j<nums.length; j++)                if(j != i)                    tmp *= nums[j];            res[i] = tmp;        }        return res;    }//两次遍历法    public int[] productExceptSelf2(int[] nums) {        int n = nums.length;        int[] res = new int[n];        res[0] = 1;        for (int i = 1; i < n; i++) {            res[i] = res[i - 1] * nums[i - 1];        }        int right = 1;        for (int i = n - 1; i >= 0; i--) {            res[i] *= right;            right *= nums[i];        }        return res;    }