35. Search Insert Position-白红宇

35. Search Insert Position

阅读量：6069 次

发布时间：2019-06-20

本文共 1076 字，大约阅读时间需要 3 分钟。

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.Example 1:Input: [1,3,5,6], 5Output: 2Example 2:Input: [1,3,5,6], 2Output: 1Example 3:Input: [1,3,5,6], 7Output: 4Example 1:Input: [1,3,5,6], 0Output: 0

解析

// 35. Search Insert Positionclass Solution_35 {public:    int searchInsert1(vector
    
     & nums, int target) {        int low = 0, high = nums.size() - 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;            }        }        return low; //没有查找到，返回的low,即插入位置    }    int searchInsert(int A[], int n, int target) {        vector
     
       vec(A, A + n);        return searchInsert1(vec, target);    }};