leetcode 33. Search in Rotated Sorted Array
Problems:
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).
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.
解题思路:
一开始我还是使用了传统的mid 与 target 进行比较的方式特别复杂
code:
public class Solution {
public int search(int[] nums, int target) {
int start = -1;
int end = nums.length;
if(nums.length==0||nums==null){
return -1;
}
int last = nums[nums.length-1];
int first = nums[0];
while(start+1<end){
int mid = start+(end-start)/2;
if(nums[mid]==target){
return mid;
}
if(nums[mid]>target){
if(first>target&&first<=nums[mid]){
start = mid;
first = nums[start];
}else{
end = mid;
last = nums[end];
}
}else{
if(last<target&&last>=nums[mid]){
end = mid;
last = nums[end];
}else{
start = mid;
first = nums[start];
}
}
}
if(start>=0&&nums[start]==target){
return start;
}
if(end<nums.length&&nums[end]==target){
return end;
}
return -1;
}
}
解题思路二:
大神的思路:
对于有序数组,使用二分搜索比较方便。分析题中的数组特点,旋转后初看是乱序数组,但仔细一看其实里面是存在两段有序数组的。刚开始做这道题时可能会去比较 target 和 A[mid] , 但分析起来异常复杂。该题较为巧妙的地方在于如何找出旋转数组中的局部有序数组,并使用二分搜索解之。结合实际数组在纸上分析较为方便。
1.若 target == A[mid] ,索引找到,直接返回
2.寻找局部有序数组,分析 A[mid] 和两段有序的数组特点,由于旋转后前面有序数组最小值都比后面有序数组最大值大。故若 A[start] < A[mid] 成立,则start与mid间的元素必有序(要么是前一段有序数组,要么是后一段有序数组,还有可能是未旋转数组)。
3.接着在有序数组 A[start]~A[mid] 间进行二分搜索,但能在 A[start]~A[mid] 间搜索的前提是 A[start] <= target <= A[mid] 。
4.接着在有序数组 A[mid]~A[end] 间进行二分搜索,注意前提条件。
5.搜索完毕时索引若不是mid或者未满足while循环条件,则测试A[start]或者A[end]是否满足条件。
6.最后若未找到满足条件的索引,则返回-1.
code:
public int search(int[] A, int target) {
if (A == null || A.length == 0) return -1;
int lb = 0, ub = A.length - 1;
while (lb + 1 < ub) {
int mid = lb + (ub - lb) / 2;
if (A[mid] == target) return mid;
if (A[mid] > A[lb]) {
// case1: numbers between lb and mid are sorted
if (A[lb] <= target && target <= A[mid]) {
ub = mid;
} else {
lb = mid;
}
} else {
// case2: numbers between mid and ub are sorted
if (A[mid] <= target && target <= A[ub]) {
lb = mid;
} else {
ub = mid;
}
}
}
if (A[lb] == target) {
return lb;
} else if (A[ub] == target) {
return ub;
}
return -1;
}