[Leetcode + Lintcode] 35. Search Insert Position

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

 

 binary-search的变种，相当于找第一个大于等于target的element

1. java

public class Solution {
    public int searchInsert(int[] nums, int target) {
        if(nums == null || nums.length == 0){
            return 0;
        }
        int start = 0;
        int end = nums.length - 1;
        while(start + 1 < end){
            int mid = start + (end - start)/2;
            if(nums[mid] < target){
                start = mid;
            }
            else if(nums[end] > target){
                end = mid;
            }
            else{
                return mid;
            }
        }
        if(nums[start] >= target){
            return start; // nums[start] is the first element >= target
        }
        else if(nums[end] >= target){
            return end; // nums[end] is the first element >= target
        }
        else{
            return end+1; //nums[end] is the last element < target, should insert behind it
        }
    }
}

2. python

class Solution:
    """
    @param A : a list of integers
    @param target : an integer to be inserted
    @return : an integer
    """
    def searchInsert(self, A, target):
        # write your code here
        if A is None or len(A) == 0:
            return 0
        
        start = 0
        end = len(A) - 1
        while(start + 1 < end):
            mid = start + (end - start)/2
            if A[mid] == target:
                return mid
            elif A[mid] < target:
                start = mid
            else:
                end = mid
        
        if A[start] >= target:
            return start
        elif A[end] >= target:
            return end
        else:
            return end+1