题目链接

题目大意:在一个旋转数组中,判断给定的target是否存在于该旋转数组中。数组中没有重复数值。例子如下:

法一:二分。确定中间元素之后,就要判断下一步是遍历左数组还是遍历右数组。如果左数组有序,且target在左数组范围内,则遍历左数组,否则遍历右数组;如果右数组有序,且target在右数组范围内,则遍历右数组,否则遍历左数组。o(lgn)。代码如下(耗时15ms):

     public int search(int[] nums, int target) {
int left = 0, right = nums.length - 1;
while(left <= right) {
int mid = (left + right) / 2;
if(nums[mid] == target) {
return mid;
}
//如果左半段有序
else if(nums[left] <= nums[mid]) {
//如果target在左半段范围内
if(nums[left] <= target && nums[mid] > target) {
right = mid - 1;
}
//虽然左半段有序,如果target不在左半段范围内,则向右遍历右半段
else {
left = mid + 1;
}
}
//如果右半段有序
else {
//如果target在右半段范围内
if(nums[right] >= target && nums[mid] < target) {
left = mid + 1;
}
//虽然右半段有序,如果target不在右半段范围内,则向左遍历左半段
else {
right = mid - 1;
}
}
}
return -1;
}

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