LeetCode 35:Search Insert Position

最新推荐文章于 2021-05-03 16:49:38 发布

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本文探讨了在已排序数组中寻找特定元素的插入位置的算法，对比了一种线性遍历方法与二分查找法，阐述了各自的时间复杂度，并通过实例展示了算法的应用。

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

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//方法一：直接遍历，时间复杂度为O(n)
 class Solution {
 public:
	 int searchInsert(vector<int>& nums, int target) {
		 int n = nums.size();
		 if (n == 0)  return NULL;
		 else if (n == 1 && nums[0] < target)    return 1;
		 else if (n == 1 && nums[0] >= target)   return 0;
		 else
		 {
			 for (int i = 0; i < n;)
			 {
				 if (nums[i] == target)
					 return i;
				 else if (target < nums[i]) 
					 return i;
				 else if (target>nums[i] && target<nums[i + 1])
					 return i + 1;
				 else
					 i++;
			 }
		 }
	 }
 };

//方法二：利用二分法查找的思想，时间复杂度为O(logn)
 class Solution {
 public:
	 int searchInsert(vector<int>& nums, int target) {
		 int low = 0, high = nums.size() - 1;

		 while (low <= high) {
			 int mid = low + (high - low) / 2;

			 if (nums[mid] < target)
				 low = mid + 1;
			 else
				 high = mid - 1;
		 }

		// (1) whilie循环结束后，low>high，即low>=high+1;
		// (2) 因为目标索引值在 [low, high+1]中, 有 low <= high+1，再根据（1）中low>=high+1，可以得到 low == high+1.
		// (3) 根据 (2), 我们可以知道目标索引值是在 [low, high+1] = [low, low], 即low就是所求的目标索引值

		 return low;
	 }
 };