假設按照升序排序的數組在預先未知的某個點上進行了旋轉。
( 例如,數組 [0,1,2,4,5,6,7] 可能變爲 [4,5,6,7,0,1,2] )。
搜索一個給定的目標值,如果數組中存在這個目標值,則返回它的索引,否則返回 -1 。
你可以假設數組中不存在重複的元素。
你的算法時間複雜度必須是 O(log n) 級別。
示例 1:
輸入: nums = [4,5,6,7,0,1,2], target = 0
輸出: 4
示例 2:
輸入: nums = [4,5,6,7,0,1,2], target = 3
輸出: -1
來源:力扣(LeetCode)
鏈接:https://leetcode-cn.com/problems/search-in-rotated-sorted-array
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class Solution {
public int search(int[] nums, int target) {
int len = nums.length;
if(len < 1) return -1;
return find(nums, 0, len-1, target);
}
private int find(int[] nums, int left, int right, int target) {
if(left == right) {
if(nums[left] == target) return 0;
return -1;
}
int mid = left + (right - left)/2;
int leftIndex = finding(nums, left, mid, target);
int rightIndex = finding(nums, mid+1, right, target);
return Math.max(leftIndex, rightIndex);
}
private int finding(int[] nums, int left, int right, int target) {
for(int i = left; i <= right; i++) {
if(target == nums[i]) {
return i;
}
}
return -1;
}
}