题目:
搜索插入位置
Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.
You may assume no duplicates in the array.
Here are few examples.
[1, 3, 5, 6], 5 -> 2
[1, 3, 5, 6], 2 -> 1
[1, 3, 5, 6], 7 -> 4
[1, 3, 5, 6], 0 -> 0
分析:
这是典型的二分搜索法。如果使用二分模版是可以直接适用的,在模板中,while循环的条件是min <= max,如果target和mid指向的数值相等,则返回mid,否则根据情况min=mid+1或者max=mid-1。这样的好处是,如果找不到该数,max是比该数小的那个数的下标,而min是比该数大的那个数的下标。这题中,我们返回min就行了,如果返回max,要注意-1的情况。
代码:
public class SearchInsertPosition {
public int searchInsertPosition(int[] nums, int target) {
int min = 0, max = nums.length - 1;
//条件是min <= max
while(min <= max) {
int mid = (min + max) / 2;
if(nums[mid] == target)//找到了,返回数组中的索引位置
return mid;
else if(nums[mid] > target)
max = mid - 1;
else
min = mid + 1;
}
return min;//未找到,返回插入位置
}
}