033 - Search in Rotated Sorted Array

Suppose a sorted array 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.

一个排好序的数列，把分割成多块，再乱序交叉，再查找一个数

int *findright(int *num, int numsize)
{
int left = 0, right = numsize - 1, mid;
while (left <= right) {
mid = (left + right) / 2;
if (num[mid] >= *num) left = mid + 1;
else if(num[mid] < *num) right = mid - 1;
}
return num + left - 1;
}

int findmid(int *nums, int *x, int size, int target)
{
int left = 0, right = size - 1, mid;
while (left <= right) {
mid = (left + right) / 2;
if (x[mid] < target) left = mid + 1;
else if(x[mid] > target) right = mid - 1;
else break;
}
return left > right? -1 : x - nums + mid;
}
int search(int* nums, int numsSize, int target)
{
int *left, *right, *move = nums;
while (1) {
left = move;
right = findright(move, numsSize - (left - nums));
if (*left > target) move = right + 1;
else if (*right < target) move = right + 1;
else {
return findmid(nums, left, right - left + 1, target);
}
if (right - nums == numsSize - 1) return -1;
}
}