[LeetCode] 81. Search in Rotated Sorted Array II

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).

Write a function to determine if a given target is in the array.

The array may contain duplicates.

题意及分析：一个升序数组在某个点被翻转过，判断target在数组中的位置，若不存在返回-1,链表中可能含重复的。如果可能有重复，那我们上一题判断左右部分是否有序的方法就失效了，因为可能有这种13111情况，虽然起点小于等于中间，但不代表右边就不是有序的，因为中点也小于等于终点，所有右边也是有序的。所以，如果遇到这种中点和两边相同的情况，我们可以直接向前移动，删除重复点。

代码

class Solution {public boolean search(int[] nums, int target) {int start = 0, end = nums.length - 1, mid = -1;while(start <= end){mid = (start + end) /2;if(target == nums[mid])return true;if(nums[mid] < nums[end] || nums[mid] < nums[start]){       //如果明确右边有序或者左边无序，可以直接在右边查找if(target > nums[mid] && target <= nums[end]){start = mid + 1;}elseend = mid -1;}else if(nums[start] < nums[mid] || nums[mid] > nums[end]){  //如果明确左边有序或者右边无序，可以直接在左边查找if(target < nums[mid] && target >=nums[start]){end = mid -1;}elsestart = mid +1;}else{      //出现num[start]==nums[mid]==nums[end],直接向前移动一个，这样可以达到删除重复数。end--;}}return false; }
}

转载于:https://www.cnblogs.com/271934Liao/p/8056392.html

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