Search in Rotated Sorted Array I II
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原题:
Suppose a sorted array 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
).
You are given a target value to search. If found in the array return its index, otherwise return -1.
You may assume no duplicate exists in the array.
解析:
没有重复元素的情形下:
- 如果中间元素大于首部元素,那么表明查找区间左半部分是有序的,然后再根据target是否在有序的一部分来决定接下来查找的方向
- 如果中间元素小于首部元素,那么查找区间有半部分是有序的,然后再根据target是否在有序的一部分来决定接下来查找的方向
- 如果中间元素等于首部元素,此时只有可能是查找区间包含一个或两个元素的情形,让左边界加1继续查找即可(相当于左边界 = middle + 1,因为此时midlle等于左边界)
包含重复元素的情形:
- 前两种情况一样,最后一种情况当中间元素等于首部元素时,查找区间有可能不再只包含一个或两个元素,此时没法判断哪一部分有序,只能将查找区间左边界加1,其实操作和上面是相同的
public int search(int[] A, int target) {if(A==null || A.length<1) {return -1;}int left = 0; int right = A.length-1;while(left <= right) {int middle = left + (right-left)/2;if(A[middle]==target) {return middle;} else if (A[left] < A[middle]) {if(A[middle]>target && A[left]<=target) {right = middle-1;} else {left = middle+1;}} else if(A[middle] < A[left]) {if(A[middle]<target && A[right]>=target) {left = middle+1;} else {right = middle-1;}} else {left++;}}return -1; }
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