二分查找进阶——循环有序数组查找再进阶——循环有序重复数组查找

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就在昨天才刚刚写了一篇关于循环有序数组查找的函数：http://blog.csdn.net/qq_33724710/article/details/51200889
今天又发现了这道题的升级版，迫不及待的拿来和大家分享（依旧来自LeetCode）：
Remove Duplicates from Sorted Array II
Follow up for "Remove Duplicates":
What if duplicates are allowed at most twice?
For example,
Given sorted array nums = [1,1,1,2,2,3],
Your function should return length = 5, with the first five elements ofnums being 1, 1, 2, 2 and3. It doesn't matter what you leave beyond the new length.
有了昨天解题的思想，这道题其实无非就是多了个重复的因素：
分析（《LeetCode-cpp》）：
允许重复元素，那么我们上一题的假设nums[mid] >= nums[first],则[first, num]递增的假设就不能成立了，比如1,3,1,1,1
如果nums[mid] >= nums[first]不能确定递增，那就把它拆分成两个条件：
1、若nums[mid] >= nums[first]，则区间[first,mid]一定递增
2、若nums[mid] == nums[first]，那就first++,往下一步看即可
接下来就让我们来分析下代码：
/*  * leetcode, Search in Rotated Sorted Array II * Follow up for "Search in Rotated Sorted Array": * What if duplicates are allowed? * Would this affect the run-time complexity? How and why? * Write a function to determine if a given target is in the array. */#define _CRT_SECURE_NO_WARNINGS#include <stdio.h>#include <stdlib.h>#define true 1#define false 0typedef int bool;bool search(int* nums, int numsSize, int target) {int first = 0;int last = numsSize - 1;/* * 与无重复的相比，不同的是，当 nums[first] <= nums[mid] 时，不能以此判断[first,mid]是递增的 * 所以当 nums[first] = nums[mid] 时，++first,跳过一个重复的数，继续循环判断， * 直到nums[first] ！= nums[mid],此时的情况就与无重复的有序数组相同了 */while (first <= last) {int mid = (last - first) / 2 + first;if (*(nums + mid) == target)return true;else if (*(nums + first) < *(nums + mid)) {if (*(nums + first) <= target && *(nums + mid) > target)last = mid - 1;elsefirst = mid + 1;}else if (*(nums + first) > *(nums + mid)) {if (*(nums + last) >= target && *(nums + mid) < target)first = mid + 1;elselast = mid - 1;}else {    //跳过一个重复的数++first;}}return false;}int main(){int arr[] = { 1, 3, 1, 1, 1 };int sz = sizeof(arr) / sizeof(arr[0]);bool ret = search(arr, sz, 1);if (ret == true)printf("true\n");elseprintf("false\n");system("pause");return 0;}
有了上一题的思路，这道题是不是也不难呢？

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