题目描述
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
思路1：
时间复杂度O(n)的方法：从头扫描数组，如果大于等于target，则返回当前index。
class Solution { public int searchInsert(int[] nums, int target) { // O(n)的方法 int index = -1; for(int i = 0; i < nums.length; i++) { if(nums[i] >= target) { index = i; break; } } if(index == -1) { index = nums.length; } return index; } }思路2：
时间复杂度O(logn)的方法：既然数组是排好序的，利用二分查找法，能使时间复杂度提高至O(logn)
class Solution { public int searchInsert(int[] nums, int target) { // O(logn)的方法 int left = 0, right = nums.length-1; while(left <= right) { int mid = (left + right) >> 1; if(nums[mid] == target) { return mid; }else if(nums[mid] < target) { left = mid + 1; }else { right = mid - 1; } } return left; } }