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Search in Rotated Sorted Array I && II
对翻过一次的排序数组二分查找，要利用好已排序这个条件
class Solution {
public:
int search(int A[], int n, int target) {
int left = 0, right = n-1;
while(left <= right){
int mid = (left+right)/2;
if(A[mid] == target)
return mid;
if(A[left] <= A[mid]){
if(A[left] <= target && target < A[mid])
right = mid-1;
else
left = mid+1;
}
else {
if(A[mid] < target && target <= A[right])
left = mid+1;
else
right = mid-1;
}
}
return -1;
}
};II中允许重复数字，碰到相等的向前移。
class Solution {
public:
bool search(int A[], int n, int target) {
int left = 0, right = n-1;
while(left <= right){
int mid = (left+right)/2;
if(A[mid] == target)
return true;
if(A[left] < A[mid]){
if(A[left] <= target && target < A[mid])
right = mid-1;
else
left = mid+1;
}
else if(A[left] > A[mid]){
if(A[mid] < target && target <= A[right])
left = mid+1;
else
right = mid-1;
}
else left++;
}
return false;
}
};Search in Rotated Sorted Array I && II
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