[LeetCode] 35. Search Insert Position 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.

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 1:

Input: [1,3,5,6], 0
Output: 0

给一个有序数组和一个目标值，数组中没有重复的元素，如果找到目标元素就返回index，如果没有找到就返回要插入的使其有序的位置。

解法1：暴力搜索Brute force

解法2：二分法Binary Search，这个显然是要考察的方法。

Java:

class Solution {
    public int searchInsert(int[] A, int target) {
        int low = 0, high = A.length-1;
        while(low<=high){
            int mid = (low+high)/2;
            if(A[mid] == target) return mid;
            else if(A[mid] > target) high = mid-1;
            else low = mid+1;
        }
        return low;
    }
}　

Java:

public int searchInsert(int[] nums, int target) {
    int low = 0, high = nums.length;
    while(low < high) {
        int mid = low + (high - low) / 2;
        if(nums[mid] < target)
            low = mid + 1;
        else
            high = mid;
    }
    return low;
}　　

Python:

class Solution(object):
    def searchInsert(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: int
        """
        left, right = 0, len(nums) - 1
        while left <= right:
            mid = left + (right - left) / 2
            if nums[mid] >= target:
                right = mid - 1
            else:
                left = mid + 1

        return left　

Python: wo

class Solution(object):
    def searchInsert(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: int
        """
        left, right = 0, len(nums) - 1
        while left <= right:
            mid = left + (right - left) / 2
            if nums[mid] > target:
                right = mid - 1
            elif nums[mid] < target:
                left = mid + 1
            else:
                return mid
            
        return left 　

C++: Brute force, TLE

class Solution {
public:
    int searchInsert(vector<int>& nums, int target) {
        for (int i = 0; i < nums.size(); ++i) {
            if (nums[i] >= target) return i;
        }
        return nums.size();
    }
};

C++: BS　　

class Solution {
public:
    int searchInsert(vector<int>& nums, int target) {
        if (nums.back() < target) return nums.size();
        int left = 0, right = nums.size() - 1;
        while (left < right) {
            int mid = left + (right - left) / 2;
            if (nums[mid] == target) return mid;
            else if (nums[mid] < target) left = mid + 1;
            else right = mid;
        }
        return right;
    }
};

　　

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