【LeetCode】Search Insert Position搜索插入位置[数组、二分法]

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.
给定一个排序数组和一个目标值，如果在数组中找到目标值则返回索引。如果没有，返回到它将会被按顺序插入的位置。
你可以假设在数组中无重复元素。

样例
[1,3,5,6]，5 → 2
[1,3,5,6]，2 → 1
[1,3,5,6]， 7 → 4
[1,3,5,6]，0 → 0

挑战
O(log(n)) time
标签
二分法 数组 排序数组

（1）Java

public class Solution {    /*     * @param A: an integer sorted array     * @param target: an integer to be inserted     * @return: An integer     */    public int searchInsert(int[] A, int target) {        if(A == null || A.length == 0) return 0;        int start,end,mid;        start = 0;        end = A.length - 1;        while(start + 1 < end){            mid = start + (end - start) / 2; // avoid Stack Overflow             if(A[mid] == target){                return mid;            }else if(A[mid] < target){                start = mid;            }else{                end = mid;            }        }//法①：find the first position >= target        //若下标 start(end)处的值>=target 则直接将target插入对应的start(end)处即可        if(A[start] >= target){            return start;        }else if(A[end] >= target){            return end;        }else{// if(A[start] < target && A[end] < target)            return end + 1;        }    }}//Version 2//法②：find the last position < target, return +1， 要特判一下target小于所有数组里面的元素public class Solution {    public int searchInsert(int[] A, int target) {        if (A == null || A.length == 0) {            return 0;        }        int start = 0;        int end = A.length - 1;        int mid;        if (target < A[0]) {            return 0;        }        // find the last number less than target        while (start + 1 < end) {            mid = start + (end - start) / 2;            if (A[mid] == target) {                return mid;            } else if (A[mid] < target) {                start = mid;            } else {                end = mid;            }        }        if (A[end] == target) {            return end;        }        if (A[end] < target) {            return end + 1;        }        if (A[start] == target) {            return start;        }        return start + 1;    }}

（2）C++

class Solution {    /**      * param A : an integer sorted array     * param target :  an integer to be inserted     * return : an integer     */public:    int searchInsert(vector<int> &A, int target) {        // find first position >= target        if (A.size() == 0) {            return 0;        }        int start = 0, end = A.size() - 1;        while (start + 1 < end) {            int mid = (end - start) / 2 + start;            if (A[mid] >= target) {                end = mid;            } else {                start = mid;            }        }        if (A[start] >= target) {            return start;        }        if (A[end] >= target) {            return end;        }        return A.size();    }};