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6 | 6 | * 难度:Medium |
7 | 7 | * 分类:Array, Binary Search |
8 | 8 | * 思路:二分查找的思路,多了一步判断,判断哪部分有序,是否在这部分中 |
9 | | - * Tips:注意边界判断,是否有等号 |
| 9 | + * Tips:注意边界判断,是否有等号。挺麻烦的,没有输入样本,很难想通边边角角 |
| 10 | + * lc81 |
10 | 11 | */ |
11 | 12 | public class lc33 { |
12 | 13 | public static void main(String[] args) { |
@@ -40,4 +41,26 @@ public static int search(int[] nums, int target) { |
40 | 41 | } |
41 | 42 | return -1; |
42 | 43 | } |
| 44 | + |
| 45 | + public int search2(int[] nums, int target) { |
| 46 | + int begin = 0, end = nums.length-1; |
| 47 | + while(begin<=end){ |
| 48 | + int mid = (begin+end)/2; |
| 49 | + if(target==nums[mid]) return mid; |
| 50 | + if(nums[begin]<=nums[mid]){ //左边有序,别忘了=,begin==end |
| 51 | + if(target>=nums[begin]&&target<nums[mid]){ |
| 52 | + end = mid-1; |
| 53 | + }else{ |
| 54 | + begin = mid+1; |
| 55 | + } |
| 56 | + }else{ //右边有序 |
| 57 | + if(target>nums[mid]&&target<=nums[end]){ //别忘了= |
| 58 | + begin = mid+1; |
| 59 | + }else{ |
| 60 | + end = mid-1; |
| 61 | + } |
| 62 | + } |
| 63 | + } |
| 64 | + return -1; |
| 65 | + } |
43 | 66 | } |
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