35. Search Insert Position
Given a sorted array of distinct integers 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 must write an algorithm with O(log n) runtime complexity.
Example 1:
Input: nums = [1,3,5,6], target = 5 Output: 2
Example 2:
Input: nums = [1,3,5,6], target = 2 Output: 1
Example 3:
Input: nums = [1,3,5,6], target = 7 Output: 4
Constraints:
1 <= nums.length <= 104-104 <= nums[i] <= 104numscontains distinct values sorted in ascending order.-104 <= target <= 104
Solution：
binary search
二分搜索
class Solution {
/* 二分搜索
*/
public int searchInsert(int[] nums, int target) {
int start =0;
int end =nums.length-1;
while(start <= end){
// 每次loop需要更新索引mid为当前数组的中间位置
int mid = start + (end-start)/2;
// int mid = (end+start)/2;
if(nums[mid] == target){
return mid;
} else if(nums[mid] > target){
end = mid - 1;
}else if(nums[mid] < target){
start = mid + 1;
}
}
return start ;
}
}
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