35.搜索插入位置

题目描述: 给定一个排序数组和一个目标值,在数组中找到目标值,并返回其索引。如果目标值不存在于数组中,返回它将会被按顺序插入的位置。
你可以假设数组中无重复元素。
示例 1:
输入: [1,3,5,6], 5
输出: 2

  • 顺序查找:时间复杂度O(n)
//C

int searchInsert(int* nums, int numsSize, int target){
    int i;
    for(i = 0; i < numsSize; i++){
        if(nums[i] >= target) return i;
    }
    return i;
}

//JS
var searchInsert = function(nums, target) {
    var i = 0, len = nums.length;

    for(; i < len; i++) {
        if(nums[i] >= target) 
            return i;        
    }
    return i;
};

  

  • 折半查找:时间复杂度O(logn)
//C

int searchInsert(int* nums, int numsSize, int target){
    int low = 0, high = numsSize - 1, mid;
    while(low <= high){
        mid = (low + high) / 2;
        if(nums[mid] == target) return mid;
        else if(nums[mid] > target) high = mid - 1;
        else low = mid + 1;
    }
    return low;
}

//JS

var searchInsert = function(nums, target) {
    let low = 0, high = nums.length - 1, mid = 0;
    while(low <= high){
        mid = Math.floor((low + high) / 2);
        if(nums[mid] == target) return mid;
        else if(nums[mid] > target) high = mid - 1;
        else low = mid + 1;
    }
    return low;
};

  

原文地址:https://www.cnblogs.com/JesseyWang/p/13083575.html