35. Search Insert Position

题目：

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

代码：

第一反应，遍历了一遍：

public int searchInsert(int[] nums, int target) {
int result =0;
for (int i = 0 ; i < nums.length ; i++)
{
if (target <= nums[i])
{
result = i;
break;
}
else
{
result=i+1;
}
}
return result;
}

网上查了一下，发现二分法，算法复杂度会减少不少，尤其在较大的数据量：

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