常用算法模板

二分法

左闭右闭 [left, right]

public int search(int[] nums, int target) {
    int l = 0, r = nums.length - 1;
    while (l <= r) {
        int mid = l + (r - l) / 2;
        if (nums[mid] == target) {
            return mid;
        } else if (nums[mid] < target) {
            l = mid + 1;
        } else {
            r = mid - 1;
        }
    }
    return -1;
}

int search(vector<int>& nums, int target) {
    int l = 0, r = nums.size() - 1;
    while (l <= r) {
        int mid = l + (r - l) / 2;
        if (nums[mid] == target) {
            return mid;
        } else if (nums[mid] < target) {
            left = mid + 1;
        } else {
            right = mid - 1;
        }
    }
    return -1;
}

左闭右开 [left, right)

public int search(int[] nums, int target) {
    int l = 0, r = nums.length - 1;
    while (l < r) {
        int mid = l + (r - l) / 2;
        if (nums[mid] == target) {
            return mid;
        } else if (nums[mid] < target) {
            l = mid + 1;
        } else {
            r = mid;
        }
    }
    return -1;
}

int search(vector<int>& nums, int target) {
    int l = 0, r = nums.size() - 1;
    while (l < r) {
        int mid = l + (r - l) / 2;
        if (nums[mid] == target) {
            return mid;
        } else if (nums[mid] < target) {
            left = mid + 1;
        } else {
            right = mid;
        }
    }
    return -1;
}

左闭右闭 or 左闭右开，记一种即可

二叉树层次遍历

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),
 * right(right) {}
 * };
 */
vector<vector<int>> levelOrder(TreeNode* root) {
    vector<vector<int>> res;
    if (root == NULL) return res;
    
    queue<TreeNode*> que;
    que.push(root);
    while (!que.empty()) {
        int size = que.size();
        vector<int> level;
        while (size-- > 0) {
            TreeNode* node = que.front();
            que.pop();
            level.push_back(node->val);
            if (node->left) que.push(node->left);
            if (node->right) que.push(node->right);
        }
        res.push_back(level);
    }
    return res;
}

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public List<List<Integer>> levelOrder(TreeNode root) {
    List<List<Integer>> res = new ArrayList<>();
    if (root == null) return res;
    
    Queue<TreeNode> que = new LinkedList<>();
    que.offer(root);
    while (!que.isEmpty()) {
        int size = que.size();
        List<Integer> level = new ArrayList<>();
        while (size-- > 0) {
            TreeNode node = que.poll();
            level.add(node.val);
            if (node.left != null) que.offer(node.left);
            if (node.right != null) que.offer(node.right);
        }
        res.add(level);
    }
    return res;
}

双指针法

移除元素

题目链接：LeetCode27.移除元素

原地移除数组 nums 中所有数值等于 val 的元素，并返回移除后数组的新长度

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

public int removeElement(int[] nums, int val) {
    int slow = 0, fast = 0;
    while (fast < nums.length) {
        if (nums[fast] != val) {
            nums[slow++] = nums[fast];
        }
        ++fast;
    }
    return slow;
}

反转字符串

题目链接：LeetCode344.反转字符串

将输入的字符串（以字符数组 s 的形式给出）原地反转过来

void reverseString(vector<char>& s) {
    for (int l = 0, r = s.size() - 1; l < r; ++l, --r) {
        swap(s[l], s[r]);
    }
}

public void reverseString(char[] s) {
    for (int l = 0, r = s.length - 1; l < r; ++l, --r) {
        s[l] ^= s[r];
        s[r] ^= s[l];
        s[l] ^= s[r];
    }
}

滑动窗口

题目链接：LeetCode209.长度最小的子数组

找出数组 nums 中满足总和大于等于 target 的长度最小的连续子数组，返回其长度

int minSubArrayLen(int target, vector<int>& nums) {
    int left = 0, right = 0;
    int sum = 0;
    int minLen = INT32_MAX;
    int curLen = 0;
    while (right < nums.size()) {
        sum += nums[right];
        while (sum >= target) {
            curLen = right - left + 1;
            if ()
            if (curLen < minLen) minLen = curLen;
            sum -= nums[left++];
        }
        ++right;
    }
    return minLen == INT32_MAX ? 0 : minLen;
}

public int minSubArrayLen(int target, int[] nums) {
    int minLen = Integer.MAX_VALUE;
    int sum = 0;
    for (int i = 0, j = 0; j < nums.length; ++j) {
        sum += nums[j];
        while (sum >= target) {
            int len = j - i + 1;
            if (len == 1) return 1;
            minLen = Math.min(minLen, len);
            sum -= nums[i++];
        }
    }
    return minLen == Integer.MAX_VALUE ? 0 : minLen;
}