LeetCode 35 Search Insert Position 二分法处理

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

解题思路：这道题目的意思是从序列中找到指定的target，返回index。若找不到，则返回插入该数的正确位置。因此，在前面整个就是基本的二分法处理，只有当找不到的时候，有r==l。分三种情况处理，左右指针在最左边界、最右边界以及中间某位置的三种情况处理。

代码如下：

 public int searchInsert(int[] nums, int target) {int l = 0;int len = nums.length;int r = len-1;if(len == 1){if(target<=nums[0]){return 0;}else{return 1;}}int mid =0;while(l<r){mid = (l+r)/2;if(target==nums[mid]){ return mid;}else if(target>nums[mid]){l = mid+1;}else{r = mid-1;}}//左边界  右边界  中间 三种情况        if(l==0){        if(target<=nums[l]){        return 0;        }else{        return 1;        }        }else if(l==len-1){          if(target>nums[l]){        return len;        }else{        return len-1;        }        }else{        if(target>nums[l]){        return l+1;        }else{        return l;        }        }      }