leetcode练习:27. Remove Element & 35. Search Insert Position

Posted 年糕君の勉强笔记

tags:

篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了leetcode练习:27. Remove Element & 35. Search Insert Position相关的知识,希望对你有一定的参考价值。

Given an array and a value, remove all instances of that value in place and return the new length.

Do not allocate extra space for another array, you must do this in place with constant memory.

The order of elements can be changed. It doesn‘t matter what you leave beyond the new length.

Example:
Given input array nums = [3,2,2,3]val = 3

Your function should return length = 2, with the first two elements of nums being 2.

var removeElement = function(nums, val) {
    var i,j,k;
    
    var len = nums.length;
    for(i=0;i<nums.length;i++){
        if(nums[i] == val) {
            nums.splice(i,1);         
            i--;
        }
    }
    return nums.length;
};

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

var searchInsert = function(nums, target) {
    var len = nums.length;
    
    if(nums[0] >= target) {
        return 0;
    }
    
    for(var i = 0; i<len-1 ; i++){
            
        if(nums[i+1] == target)
            return i+1;
        
        if(nums[i] < target && nums[i+1] > target){
            return i+1;
        }
    }
    
    if(i==len-1){
        return nums.length;
    }
};

 







以上是关于leetcode练习:27. Remove Element & 35. Search Insert Position的主要内容,如果未能解决你的问题,请参考以下文章

leetcode 27 Remove Element

[Leetcode] 27 Remove Element

LeetCode OJ 27. Remove Element

#Leetcode# 27. Remove Element

[LeetCode]27. Remove Element

LeetCode 27 Remove Element