LeetCode--Array--Remove Element && Search Insert Position(Easy)

27. Remove Element (Easy)# 2019.7.7

Given an array nums and a value val, 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 by modifying the input array in-place with O(1) extra memory.

The order of elements can be changed. It doesn't matter what you leave beyond the new length.

Example 1:

Given nums = [3,2,2,3], val = 3,

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

It doesn't matter what you leave beyond the returned length.

Example 2:

Given nums = [0,1,2,2,3,0,4,2], val = 2,

Your function should return length = 5, with the first five elements of nums containing 0, 1, 3, 0, and 4.

Note that the order of those five elements can be arbitrary.

It doesn't matter what values are set beyond the returned length.

solution##

class Solution {
    public int removeElement(int[] nums, int val) {
        int newlength = 0;
        for (int i = 0 ; i < nums.length; i++)
        {
            if (nums[i] != val)
                nums[newlength++] = nums[i];
        }
        return newlength;
    }
}

总结##

此题很水，没什么技术含量，不做过多解释！!

35. Search Insert Position (Easy)#2019.7.7

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], 5
Output: 2
Example 2:

Input: [1,3,5,6], 2
Output: 1
Example 3:

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

Input: [1,3,5,6], 0
Output: 0

solution##

class Solution {
    public int searchInsert(int[] nums, int target) {
        for (int i = 0; i < nums.length; i++)
        {
            if (nums[i] >= target)
                return i;
        }
        return nums.length;
    }
}

总结##

此题很水，没什么技术含量，同样不做过多解释！!