【leetcode】33. Search in Rotated Sorted Array【java】
2016-12-23 11:14
537 查看
【题目】
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.
【二分思路】
分情况讨论,数组可能有以下三种情况:
![](https://oscdn.geek-share.com/Uploads/Images/Content/201612/509a67fdcf597bbe9e5fee2def84819a)
然后,再看每一种情况中,target在左边还是在右边,其中第一种情况还可以直接判断target有可能不在数组范围内。
【Java代码】
[java] view
plain copy
![](https://oscdn.geek-share.com/Uploads/Images/Content/201611/a7c8e286f463007e2a900848b93dd72c.png)
![](https://oscdn.geek-share.com/Uploads/Images/Content/201611/9e12f1d3e499fc949c886e7c9e0484f9)
public class Solution {
public int search(int[] A, int target) {
int len = A.length;
if (len == 0) return -1;
return binarySearch(A, 0, len-1, target);
}
public int binarySearch(int[] A, int left, int right, int target) {
if (left > right) return -1;
int mid = (left + right) / 2;
if (A[left] == target) return left;
if (A[mid] == target) return mid;
if (A[right] == target) return right;
//图示情况一
if (A[left] < A[right]) {
if (target < A[left] || target > A[right]) { //target不在数组范围内
return -1;
} else if (target < A[mid]) { //target在左边
return binarySearch(A, left+1, mid-1, target);
} else { //target在右边
return binarySearch(A, mid+1, right-1, target);
}
}
//图示情况二
else if (A[left] < A[mid]) {
if (target > A[left] && target < A[mid]) { //target在左边
return binarySearch(A, left+1, mid-1, target);
} else { //target在右边
return binarySearch(A, mid+1, right-1, target);
}
}
//图示情况三
else {
if (target > A[mid] && target < A[right]) { //target在右边
return binarySearch(A, mid+1, right-1, target);
} else{ //target在左边
return binarySearch(A, left+1, mid-1, target);
}
}
}
}
下面是参考网上的思路,其中if (target == A[left]) 和 if (target == A[right]) 两个判断是我自己加上的,因为加上这两个判断后,下面分情况讨论时就不用考虑target等于边界的情况了。
[java] view
plain copy
![](https://oscdn.geek-share.com/Uploads/Images/Content/201611/a7c8e286f463007e2a900848b93dd72c.png)
![](https://oscdn.geek-share.com/Uploads/Images/Content/201611/9e12f1d3e499fc949c886e7c9e0484f9)
public class Solution {
public int search(int[] A, int target) {
int len = A.length;
if (len == 0) {
return -1;
} else if (len == 1) {
return target==A[0] ? 0 : -1;
}
int left = 0, right = len-1;
while (left < right) {
int mid = left + (right-left)/2;
if (target == A[mid]) {
return mid;
} else if (target == A[left]) {
return left;
} else if (target == A[right]) {
return right;
}
//第一种情况中,target不在数组范围内
if (A[left]<A[right] && (target<A[left] || target>A[right])) {
return -1;
}
//第一、二种情况的左边,即连续上升的左边,且target在这段内
if (A[left]<A[mid] && target>A[left] && target<A[mid]) {
right = mid - 1;
continue;
}
//第一、三种情况的右边,即连续上升的右边,且target在这段内
if (A[mid]<A[right] && target>A[mid] && target<A[right]) {
left = mid + 1;
continue;
}
//如果上面情况都不满足,那么可能在第二种情况的右边
if (A[mid] > A[right]) {
left = mid + 1;
continue;
}
//如果上面情况都不满足,第三种情况的左边
if (A[left] > A[mid]) {
right = mid - 1;
continue;
}
}
return -1;
}
}
个人感觉还是自己那样写思路比较清晰。
2015/3/30更新
[java] view
plain copy
![](https://oscdn.geek-share.com/Uploads/Images/Content/201611/a7c8e286f463007e2a900848b93dd72c.png)
![](https://oscdn.geek-share.com/Uploads/Images/Content/201611/9e12f1d3e499fc949c886e7c9e0484f9)
public class Solution {
public int search(int[] A, int target) {
int l = 0;
int r = A.length - 1;
while (l <= r) {
int mid = (l + r) / 2;
if (target == A[mid]) return mid;
if (A[l] <= A[r]) {
if (target < A[mid]) r = mid - 1;
else l = mid + 1;
} else if (A[l] <= A[mid]) {
if (target > A[mid] || target < A[l]) l = mid + 1;
else r = mid - 1;
} else {
if (target < A[mid] || target > A[r]) r = mid - 1;
else l = mid + 1;
}
}
return -1;
}
}
我自己的代码 2016-12-23
public class Solution {
public int search(int[] nums, int target) {
int minIdx = findMinIdx(nums);
if (target == nums[minIdx]){
return minIdx;
}
int m = nums.length;
int start = (target <= nums[m - 1]) ? minIdx : 0;
int end = (target > nums[m - 1]) ? minIdx : m -1;
while (start <= end){
int mid = start + (end - start) / 2;
if (nums[mid] == target){
return mid;
} else if (target > nums[mid]){
start = mid + 1;
} else {
end = mid -1;
}
}
return -1;
}
public int findMinIdx(int[] nums){
int start = 0;
int end = nums.length -1;
while (start < end){
int mid = start + (end - start) / 2;
if (nums[mid] > nums[end]){
start = mid + 1;
} else {
end = mid;
}
}
return start;
}
}
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.
【二分思路】
分情况讨论,数组可能有以下三种情况:
然后,再看每一种情况中,target在左边还是在右边,其中第一种情况还可以直接判断target有可能不在数组范围内。
【Java代码】
[java] view
plain copy
![](https://oscdn.geek-share.com/Uploads/Images/Content/201611/a7c8e286f463007e2a900848b93dd72c.png)
public class Solution {
public int search(int[] A, int target) {
int len = A.length;
if (len == 0) return -1;
return binarySearch(A, 0, len-1, target);
}
public int binarySearch(int[] A, int left, int right, int target) {
if (left > right) return -1;
int mid = (left + right) / 2;
if (A[left] == target) return left;
if (A[mid] == target) return mid;
if (A[right] == target) return right;
//图示情况一
if (A[left] < A[right]) {
if (target < A[left] || target > A[right]) { //target不在数组范围内
return -1;
} else if (target < A[mid]) { //target在左边
return binarySearch(A, left+1, mid-1, target);
} else { //target在右边
return binarySearch(A, mid+1, right-1, target);
}
}
//图示情况二
else if (A[left] < A[mid]) {
if (target > A[left] && target < A[mid]) { //target在左边
return binarySearch(A, left+1, mid-1, target);
} else { //target在右边
return binarySearch(A, mid+1, right-1, target);
}
}
//图示情况三
else {
if (target > A[mid] && target < A[right]) { //target在右边
return binarySearch(A, mid+1, right-1, target);
} else{ //target在左边
return binarySearch(A, left+1, mid-1, target);
}
}
}
}
下面是参考网上的思路,其中if (target == A[left]) 和 if (target == A[right]) 两个判断是我自己加上的,因为加上这两个判断后,下面分情况讨论时就不用考虑target等于边界的情况了。
[java] view
plain copy
![](https://oscdn.geek-share.com/Uploads/Images/Content/201611/a7c8e286f463007e2a900848b93dd72c.png)
public class Solution {
public int search(int[] A, int target) {
int len = A.length;
if (len == 0) {
return -1;
} else if (len == 1) {
return target==A[0] ? 0 : -1;
}
int left = 0, right = len-1;
while (left < right) {
int mid = left + (right-left)/2;
if (target == A[mid]) {
return mid;
} else if (target == A[left]) {
return left;
} else if (target == A[right]) {
return right;
}
//第一种情况中,target不在数组范围内
if (A[left]<A[right] && (target<A[left] || target>A[right])) {
return -1;
}
//第一、二种情况的左边,即连续上升的左边,且target在这段内
if (A[left]<A[mid] && target>A[left] && target<A[mid]) {
right = mid - 1;
continue;
}
//第一、三种情况的右边,即连续上升的右边,且target在这段内
if (A[mid]<A[right] && target>A[mid] && target<A[right]) {
left = mid + 1;
continue;
}
//如果上面情况都不满足,那么可能在第二种情况的右边
if (A[mid] > A[right]) {
left = mid + 1;
continue;
}
//如果上面情况都不满足,第三种情况的左边
if (A[left] > A[mid]) {
right = mid - 1;
continue;
}
}
return -1;
}
}
个人感觉还是自己那样写思路比较清晰。
2015/3/30更新
[java] view
plain copy
![](https://oscdn.geek-share.com/Uploads/Images/Content/201611/a7c8e286f463007e2a900848b93dd72c.png)
public class Solution {
public int search(int[] A, int target) {
int l = 0;
int r = A.length - 1;
while (l <= r) {
int mid = (l + r) / 2;
if (target == A[mid]) return mid;
if (A[l] <= A[r]) {
if (target < A[mid]) r = mid - 1;
else l = mid + 1;
} else if (A[l] <= A[mid]) {
if (target > A[mid] || target < A[l]) l = mid + 1;
else r = mid - 1;
} else {
if (target < A[mid] || target > A[r]) r = mid - 1;
else l = mid + 1;
}
}
return -1;
}
}
我自己的代码 2016-12-23
public class Solution {
public int search(int[] nums, int target) {
int minIdx = findMinIdx(nums);
if (target == nums[minIdx]){
return minIdx;
}
int m = nums.length;
int start = (target <= nums[m - 1]) ? minIdx : 0;
int end = (target > nums[m - 1]) ? minIdx : m -1;
while (start <= end){
int mid = start + (end - start) / 2;
if (nums[mid] == target){
return mid;
} else if (target > nums[mid]){
start = mid + 1;
} else {
end = mid -1;
}
}
return -1;
}
public int findMinIdx(int[] nums){
int start = 0;
int end = nums.length -1;
while (start < end){
int mid = start + (end - start) / 2;
if (nums[mid] > nums[end]){
start = mid + 1;
} else {
end = mid;
}
}
return start;
}
}
相关文章推荐
- (Java)LeetCode-33. Search in Rotated Sorted Array
- 33.search in rotated sorted array leetcode java
- Java [leetcode 33]Search in Rotated Sorted Array
- LeetCode 33 Search in Rotated Sorted Array (C,C++,Java,Python)
- leetcode 33. Search in Rotated Sorted Array __java
- LeetCode 33 — Search in Rotated Sorted Array(C++ Java Python)
- [leetcode]33. Search in Rotated Sorted Array(Java实现)
- Leetcode 33. Search in Rotated Sorted Array (Hard) (java)
- LeetCode 33 Search in Rotated Sorted Array 二叉查找(三)
- leetcode 33 -- Search in Rotated Sorted Array
- [leetcode 33] Search in Rotated Sorted Array
- [Leetcode 33, Hard] Search in Rotated Sorted Array
- LeetCode_33---Search in Rotated Sorted Array
- [leetcode 33]Search in Rotated Sorted Array
- Search in Rotated Sorted Array leetcode java
- Search in Rotated Sorted Array II leetcode java
- [Leetcode] 33. Search in Rotated Sorted Array
- LeetCode --- 33. Search in Rotated Sorted Array
- Java for LeetCode 033 Search in Rotated Sorted Array
- LeetCode(33)Search in Rotated Sorted Array