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

分析：本题考查的是查找与插入。因为数组已经是有序的，所以可以使用插入和折半插入的思想来解决问题。

代码：

解法一：

public class Solution {    public int searchInsert(int[] nums, int target) {        // 解法一        int i;        for(i = 0; i < nums.length; i++){            if(nums[i] < target) continue;// 如果数组中对应的数据小于target，继续学习            return i;// 如果等于target或者第一个大于target，则返回当前索引值        }        return i; // 如果循环中未能找到大于target，则插入数组的最后}

解法二：（二分查找）

（递归）

public class Solution {    public int searchInsert(int[] nums, int target) {                //解法二        if(nums.length == 0 || nums == null) return 0;        return binSearchInsert(nums, target, 0, nums.length - 1);    }    public int binSearchInsert(int[] nums, int target, int start, int end){        int mid = start + (end - start) / 2;        if(target < nums[mid]){            return (start < mid) ? binSearchInsert(nums, target, start, mid - 1) : mid;// 如果start == mid，        }        if(target > nums[mid]){            return end > mid ? binSearchInsert(nums, target, mid + 1, end) : mid+1;//         }        return mid;// 如果相等就返回此值    }}

（循环）

public class Solution {    public int searchInsert(int[] nums, int target) {        int left = 0, right = nums.length - 1;        int mid;        while(left <= <span style="font-family: Arial, Helvetica, sans-serif;">right</span>){            mid = left + (right - l) / 2;            if(target == nums[mid]) return mid;            if(target < nums[mid]){                right = mid - 1;            }            if(target > nums[mid]){                left = mid + 1;            }        }        return left;        }}