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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.
思路：真题思路和http://www.cnblogs.com/vincently/p/4122528.html类似。只是如果有重复的话就不能依靠与边界的比较确定递增的序列。例如，[1,3,1,1,1]，对于A[m]>=A[l]来说，[l,m]的假设就不成立。但是还是可以观察到，左边子数组的值都要大于等于右边子数组的值。将A[m]>=A[l]拆分为两种情况：A[m]>A[l],则区间[l,m]一定递增；如果 A[m]==A[l]确定不了，那就l++，往下看一步。
时间复杂度改变为O(n),空间复杂度O(1)
1 class Solution { 2 public: 3 bool search(int A[], int n, int target) { 4 int first = 0, last = n - 1; 5 while (first <= last) { 6 const int mid = (first + last) / 2; 7 if (A[mid] == target) return true; 8 if (A[first] < A[mid]) { 9 if (A[first] <= target && target < A[mid]) 10 last = mid - 1; 11 else 12 first = mid + 1; 13 } else if (A[first] > A[mid]){ 14 if (A[mid] < target && target <= A[last]) 15 first = mid + 1; 16 else 17 last = mid - 1; 18 } else { 19 first++; 20 } 21 } 22 return false; 23 } 24 };
[LeetCode] Search in Rotated Array II
标签：style blog http io ar color sp for on
原文地址：http://www.cnblogs.com/vincently/p/4122676.html
