Half and Half 类型题
二分法的精髓在于判断目标值在前半区间还是后半区间，Half and Half类型难点在不能一次判断，可能需要一次以上的判断条件。
Maximum Number in Mountain Sequence
Given a mountain sequence of n integers which increase firstly and then decrease, find the mountain top.
样例 Given
nums = [1, 2, 4, 8, 6, 3] return 8 Given nums = [10, 9, 8, 7], return 10public int mountainSequence(int[] nums) {
// write your code here
if(nums == null || nums.length == 0){
return -1;
}
int start = 0;
int end = nums.length - 1;
while(start + 1 < end){
int mid = start + (end - start)/2;
if(nums[start] < nums[mid]){
if(nums[mid+1]<nums[mid]){
end = mid;
}
else{
start = mid;
}
}
else{
if(nums[mid-1]<nums[mid]){
start = mid;
}
else{
end = mid;
}
}
}
if(nums[start] > nums[end]){
return nums[start];
}
else{
return nums[end];
}
//return -1;
}假设有一个排序的按未知的旋转轴旋转的数组(比如，0 1 2 4 5 6 7 可能成为4 5 6 7 0 1 2)。给定一个目标值进行搜索，如果在数组中找到目标值返回数组中的索引位置，否则返回-1。你可以假设数组中不存在重复的元素。
样例
给出[4, 5, 1, 2, 3]和target=1，返回 2
给出[4, 5, 1, 2, 3]和target=0，返回 -1
思路：判断目标值是否在某一区间／跨区间，再比较目标值
public int search(int[] A, int target) {
// write your code here
if(A == null | A.length == 0){
return -1;
}
int start = 0;
int end = A.length - 1;
while(start + 1 < end){
int mid = start + (end - start)/2;
if (A[start] < A[mid]){
if(target >= A[start] && target <= A[mid]){
end = mid;
}
else{
start = mid;
}
}
else{
if(target >= A[mid] && target <= A[end]){
start = mid;
}
else{
end = mid;
}
}
}
if(A[start] == target){
return start;
}
if(A[end] == target){
return end;
}
return -1;
}