LeetCode - Search In Rotated Sorted Array
2015-03-14 00:08
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https://leetcode.com/problems/search-in-rotated-sorted-array/
Suppose a sorted array is rotated at some pivot unknown to you beforehand.
(i.e.,
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.
这道题看起来挺简单,想了半天思路也不清晰,最后还是看了别人的思路才想清楚。。。唉,,感觉自己理清各种情况的能力还是挺差啊
总之就是,要么两边都没bump,要么有一边有bump,即中间大,两边小的情况,因此,检查没有bump那边,如果target在没有bump那边的range内,就检查另一边。
bump的决定条件是 这个范围的第一个数大于最后一个数,即违反了从小到大的顺序。
Suppose a sorted array is rotated at some pivot unknown to you beforehand.
(i.e.,
0 1 2 4 5 6 7might 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.
这道题看起来挺简单,想了半天思路也不清晰,最后还是看了别人的思路才想清楚。。。唉,,感觉自己理清各种情况的能力还是挺差啊
总之就是,要么两边都没bump,要么有一边有bump,即中间大,两边小的情况,因此,检查没有bump那边,如果target在没有bump那边的range内,就检查另一边。
bump的决定条件是 这个范围的第一个数大于最后一个数,即违反了从小到大的顺序。
public int search(int[] A, int target) { int left = 0; int right = A.length-1; while(left<=right){ int mid = (left+right)/2; if(A[mid]==target) return mid; if(A[mid]>=A[left]){ //注意这里必须有=号,这个条件失败就是在暗示左边没有bump,而左边只有一个数的时候没有bump,但A[mid]=A[left] if(A[left]<=target && A[mid]>target){ right = mid-1; } else{ left = mid+1; } } else{ if(A[mid]<target && A[right]>=target){ left=mid+1; } else{ right = mid-1; } } } return -1; }
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