題目
Follow up for "Search in Rotated Sorted Array":
What if duplicates are allowed?
Would this affect the run-time complexity? How and why?
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).
Write a function to determine if a given target is in the array.
The array may contain duplicates.
解題之法
class Solution {
public:
bool search(vector<int>& nums, int target) {
int n = nums.size();
if (n == 0) return false;
int left = 0, right = n - 1;
while (left <= right) {
int mid = (left + right) / 2;
if (nums[mid] == target) return true;
else if (nums[mid] < nums[right]) {
if (nums[mid] < target && nums[right] >= target) left = mid + 1;
else right = mid - 1;
} else if (nums[mid] > nums[right]){
if (nums[left] <= target && nums[mid] > target) right = mid - 1;
else left = mid + 1;
} else --right;
}
return false;
}
};
分析
這道是之前那道 Search in Rotated Sorted Array 在旋轉(zhuǎn)有序數(shù)組中搜索 的延伸赐劣，現(xiàn)在數(shù)組中允許出現(xiàn)重復(fù)數(shù)字箱硕，這個也會影響我們選擇哪半邊繼續(xù)搜索，由于之前那道題不存在相同值，我們在比較中間值和最右值時就完全符合之前所說的規(guī)律：如果中間的數(shù)小于最右邊的數(shù)，則右半段是有序的，若中間數(shù)大于最右邊數(shù)胸蛛，則左半段是有序的。而如果可以有重復(fù)值樱报，就會出現(xiàn)來面兩種情況葬项，[3 1 1] 和 [1 1 3 1]，對于這兩種情況中間值等于最右值時迹蛤，目標值3既可以在左邊又可以在右邊民珍，那怎么辦么，對于這種情況其實處理非常簡單盗飒，只要把最右值向左一位即可繼續(xù)循環(huán)嚷量，如果還相同則繼續(xù)移，直到移到不同值為止逆趣，然后其他部分還采用 Search in Rotated Sorted Array 在旋轉(zhuǎn)有序數(shù)組中搜索 中的方法蝶溶。
