Paste_Image.png
My code:
public class Solution {
public boolean search(int[] nums, int target) {
if (nums == null || nums.length == 0)
return false;
return search(0, nums.length - 1, nums, target);
}
private boolean search(int begin, int end, int[] nums, int target) {
if (begin > end)
return false;
else {
int mid = (begin + end) / 2;
if (nums[mid] > nums[end]) { // left is sorted
if (nums[mid] < target) {
int i = mid + 1;
while (i <= end && nums[i] == nums[i - 1])
i++;
if (i >= nums.length)
return false;
else
return search(i, end, nums, target);
}
else if (nums[mid] > target) {
if (nums[begin] > target) { // in the right
int i = mid + 1;
while (i <= end && nums[i] == nums[i - 1])
i++;
if (i >= nums.length)
return false;
else
return search(i, end, nums, target);
}
else if (nums[begin] == target)
return true;
else { // in the left
int i = mid - 1;
while (i >= begin && nums[i] == nums[i + 1])
i--;
if (i < 0)
return false;
else
return search(begin + 1, i, nums, target);
}
}
else
return true;
}
else if (nums[mid] < nums[end]) { // right is sorted
if (nums[mid] > target) {
int i = mid - 1;
while (i >= begin && nums[i] == nums[i + 1])
i--;
if (i < 0)
return false;
else
return search(begin, i, nums, target);
}
else if (nums[mid] < target) {
if (target < nums[end]) {
int i = mid + 1;
while (i <= end && nums[i] == nums[i - 1])
i++;
if (i >= nums.length)
return false;
else
return search(i, end, nums, target);
}
else if (target > nums[end]) {
int i = mid - 1;
while (i >= begin && nums[i] == nums[i + 1])
i--;
if (i < 0)
return false;
else
return search(begin, i, nums, target);
}
else
return true;
}
else
return true;
}
else
if (nums[mid] == target)
return true;
else
return search(begin, end - 1, nums, target);
}
}
}
My test result：
这道题目还是和前面的差不多。然后就是要用while循环把重复的数字跳到。
然后就是同样的注意点，当nums[mid] = nums[end] 时，可能出现两种情况，右侧已经排序好或者左侧已经排序好。所以不能确定。
于是，就直接将end - 1，重新进行搜索。
所以，如果数组数字都是一样的，需要进行 O(n)复杂度，而不再是之前的binary search 的 O(log n)
**
总结： Array， binary search
**
Anyway, Good luck, Richardo!
My code:
public class Solution {
public boolean search(int[] nums, int target) {
if (nums == null || nums.length == 0)
return false;
int begin = 0;
int end = nums.length - 1;
while (begin <= end) {
int middle = (begin + end) / 2;
if (nums[middle] < nums[end]) { // right part is sorted
if (target < nums[middle]) {
end = middle - 1;
}
else if (target == nums[middle]) {
return true;
}
else if (target <= nums[end]) {
begin = middle + 1;
}
else {
end = middle - 1;
}
}
else if (nums[middle] > nums[end]) { // left part is sorted
if (target > nums[middle]) {
begin = middle + 1;
}
else if (target == nums[middle]) {
return true;
}
else if (target >= nums[0]) {
end = middle - 1;
}
else {
begin = middle + 1;
}
}
else {
if (target == nums[middle])
return true;
else
end = end - 1;
}
}
return false;
}
}
差不多的题目。就把之前的情况细分。
nums[middle] < nums[end]
nums[middle] > nums[end]
nums[middle] == nums[end] -> end - 1
差不多。然后因为出现了重复元素所以可以滑动，用while循环加速。
但是我懒得写了。第一次的版本是写了的。
Anyway, Good luck, Richardo!
提供一种新的思路就是直接traverse， 复杂度是 O(n)
上面的做法， time complexity: best: O(log n), worst: O(n)
Anyway, Good luck, Richardo! -- 08/12/2016
My code:
public class Solution {
public boolean search(int[] nums, int target) {
if (nums == null || nums.length == 0) {
return false;
}
int begin = 0;
int end = nums.length - 1;
while (begin <= end) {
int mid = begin + (end - begin) / 2;
if (target == nums[mid]) {
return true;
}
else if (nums[begin] < nums[mid]) { // left is sorted
if (nums[begin] <= target && target < nums[mid]) {
end = mid - 1;
}
else {
begin = mid + 1;
}
}
else if (nums[begin] > nums[mid]) { // right is sorted
if (nums[mid] < target && target <= nums[end]) {
begin = mid + 1;
}
else {
end = mid - 1;
}
}
else {
begin++;
}
}
return false;
}
}
much simpler code
首先判断，左右那个部分是 sorted ，然后再进行下一步。
Anyway, Good luck, Richardo! -- 09/21/2016
