数据结构系统刷题

本文为系统刷leetcode的记录，会记录自己根据代码随想录刷过的leetcode，方便直接点开刷题，时常更新
时间复杂度简记为s
空间复杂度简记为k

数组

704 二分查找
一维二分查找
（1）[left, right]

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

s:$O(logn)$
k:$O(1)$
（2）[left, right)

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

s:$O(logn)$
k:$O(1)$
二维二分查找：74. 搜索二维矩阵

class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        int m = matrix.size();
        int n = matrix[0].size();
        int low = 0;
        int high = m * n - 1;
        while (low <= high) {
            int mid = (low + high) / 2;
            int num = matrix[mid / n][mid % n]; // 第一个是确定第几行，第二个是确定第几列，相当于把matrix降维成一维，比如要找一个4*4数组的第13个元素，13/4 = 3，为第四行（行索引是0开始），13%4=1，即第四行第一个
            if (num < target) {
                low = mid + 1;
            } else if (num > target) {
                high = mid - 1;
            } else return true;
        }
        return false;
    }
};

27. 移除元素

class Solution {
public:
    int removeElement(vector<int>& nums, int val) {
        int slow = 0;
        for (int fast = 0; fast < nums.size(); fast++) {
            if (nums[fast] != val) {
                nums[slow++] = nums[fast];
            }
        }
        return slow;
    }
};

s：$O(n)$
k：$O(1)$

977. 有序数组的平方

class Solution {
public:
    vector<int> sortedSquares(vector<int>& nums) {
        int k = nums.size() - 1;
        vector<int> result(nums.size(), 0);
        for (int i = 0, j = nums.size() - 1; i <= j;) {
            if (nums[i] * nums[i] > nums[j] * nums[j]) {
                result[k--] = nums[i] * nums[i];
                i++;
            } else {
                result[k--] = nums[j] * nums[j];
                j--;
            }
        }
        return result;
    }
};

s：$O(n)$
k：$O(n)$
209. 长度最小的子数组

59. 螺旋矩阵 II