LeetCode 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

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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;
	 }
 };