算法练习Day1 | 数组

704. 二分查找

拓展
35.搜索插入位置
34.在排序数组中查找元素的第一个和最后一个位置

问题

Given an array of integers nums which is sorted in ascending order, and an integer target, write a function to search target in nums. If target exists, then return its index. Otherwise, return -1.
You must write an algorithm with O(log n) runtime complexity.

LC链接
leetcode.com/problems/bi…

文章讲解
programmercarl.com/0704.%E4%BA…

使用需满足

• 数组递增或递减
• 无重复元素

重点关注

二分法需注意边界问题, 在代码上主要体现在right的初始值, right值的更新和终止条件.

	right初始值	right值的更新	终止条件
左闭右开 [)	nums.size()	mid	left < right
左闭右闭 []	nums.size() - 1	mid - 1	left <= right

代码

• 左闭右闭

    int search(vector<int>& nums, int target) {
        int left = 0;
        int right = nums.size() - 1;

        while (left <= right) {
            int mid = (left + right) / 2;

            if (nums[mid] == target) {
                return mid;
            } else if (nums[mid] < target) {
                left = mid + 1;
            } else if (nums[mid] > target) {
                right = mid - 1;
            }
        }
        return -1;
    }

• 左闭右开

    int search(vector<int>& nums, int target) {
        int left = 0;
        int right = nums.size();

        while (left < right) {
            int mid = (left + right) / 2;
            if (nums[mid] < target) {
                left = mid + 1;
            } else if (nums[mid] > target) {
                right = mid;
            } else {
                return mid;
            }
        }

        return -1;
    }

27. 移除元素

问题

Given an integer array nums and an integer val, remove all occurrences of val in nums in-place. The order of the elements may be changed. Then return the number of elements in nums which are not equal to val.

LC链接
leetcode.com/problems/re…

文章讲解
programmercarl.com/0027.%E7%A7…

方法一: Brute Force
需要关注size值的更改方式

    int removeElement(vector<int>& nums, int val) {
        int size = nums.size();

        for (int i = 0; i < size; i++) {
            if (nums[i] == val) {
                for (int j = i; j < size - 1; j++) {
                    nums[j] = nums[j+1];
                }
                i--;
                size--;
            }
        }
        return size;
    }

方法二: Two pointers
第一个方法比较直观
第二个方法比较简洁快速

    int removeElement(vector<int>& nums, int val) {
        // for (int i = 0; i < nums.size(); i++) {
        //     if (nums[i] == val) {
        //         fast++;
        //     } else {
        //         if (fast < nums.size()) {
        //             nums[slow] = nums[fast];
        //             slow++;
        //             fast++;
        //         }
        //     }
        // }

        for (int fast = 0; fast < nums.size(); fast++) {
            if (nums[fast] != val) {
                nums[slow++] = nums[fast];
            }
        }

        return slow;
    }