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.
Example 1:
Input: [1,3,5,6], 5 Output: 2
Example 2:
Input: [1,3,5,6], 2 Output: 1
Example 3:
Input: [1,3,5,6], 7 Output: 4
Example 4:
Input: [1,3,5,6], 0 Output: 0
class Solution(object):
def searchInsert(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: int
"""
if target < nums[0]:
return 0
if target > nums[-1]:
return len(nums)
i, j = 0, len(nums)-1
while i<=j:
mid = (i+j) >> 1 # /2
if nums[mid] == target:
return mid
elif nums[mid] > target:
j = mid-1
else:
i = mid+1
return i
经典二分查找！如果没有找到则nums[j] < target < nums[i]. 因此，返回i
如果前面不加判断：
public int searchInsert(int[] nums, int target) {
if(nums.length == 0) return 0;
int mid = 0;
int lo= 0,hi=nums.length-1;
while(lo<=hi){
mid = (lo+hi)/2;
if(nums[mid] == target) return mid;
else if(target > nums[mid]) lo = mid+1;
else hi = mid-1;
}
return target < nums[mid] ? mid : mid+1;
}
leetcode 35. Search Insert Position
原文：https://www.cnblogs.com/bonelee/p/9108503.html