双指针技巧总结

一个快指针走两步,一个慢指针走一步,常用于链表问题。

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bool hasCycle(ListNode* head) {
    ListNode *slow = head, *fast = head;
    while (fast && fast->next) {
        slow = slow->next;
        fast = fast->next->next;
        if (slow == fast) return true;
    }
    return false;
}

相遇后,一个从头开始,一个从相遇点开始,同速走,再次相遇就是入口。

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ListNode* detectCycle(ListNode* head) {
    ListNode *slow = head, *fast = head;
    while (fast && fast->next) {
        slow = slow->next;
        fast = fast->next->next;
        if (slow == fast) {
            slow = head;
            while (slow != fast) {
                slow = slow->next;
                fast = fast->next;
            }
            return slow;
        }
    }
    return nullptr;
}
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ListNode* middleNode(ListNode* head) {
    ListNode *slow = head, *fast = head;
    while (fast && fast->next) {
        slow = slow->next;
        fast = fast->next->next;
    }
    return slow;
}

快指针先走 k 步,然后同速走,快到末尾时慢就是倒数第 k 个。

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ListNode* getKthFromEnd(ListNode* head, int k) {
    ListNode *fast = head, *slow = head;
    for (int i = 0; i < k; i++) fast = fast->next;
    while (fast) {
        fast = fast->next;
        slow = slow->next;
    }
    return slow;
}

一个在头,一个在尾,往中间走。

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vector<int> twoSum(vector<int>& numbers, int target) {
    int left = 0, right = numbers.size() - 1;
    while (left < right) {
        int sum = numbers[left] + numbers[right];
        if (sum == target) return {left + 1, right + 1};
        else if (sum < target) left++;
        else right--;
    }
    return {-1, -1};
}
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int maxArea(vector<int>& height) {
    int left = 0, right = height.size() - 1;
    int ans = 0;
    while (left < right) {
        int area = min(height[left], height[right]) * (right - left);
        ans = max(ans, area);
        // 移动矮的那边才有可能更大
        if (height[left] < height[right]) left++;
        else right--;
    }
    return ans;
}

排序 + 固定一个数 + 双指针找另外两个:

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vector<vector<int>> threeSum(vector<int>& nums) {
    sort(nums.begin(), nums.end());
    vector<vector<int>> ans;

    for (int i = 0; i < nums.size(); i++) {
        if (i > 0 && nums[i] == nums[i-1]) continue;  // 去重
        int left = i + 1, right = nums.size() - 1;
        while (left < right) {
            int sum = nums[i] + nums[left] + nums[right];
            if (sum == 0) {
                ans.push_back({nums[i], nums[left], nums[right]});
                while (left < right && nums[left] == nums[left+1]) left++;   // 去重
                while (left < right && nums[right] == nums[right-1]) right--; // 去重
                left++; right--;
            } else if (sum < 0) left++;
            else right--;
        }
    }
    return ans;
}
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bool isPalindrome(string s) {
    int left = 0, right = s.size() - 1;
    while (left < right) {
        if (s[left] != s[right]) return false;
        left++; right--;
    }
    return true;
}

本质也是双指针,left 和 right 维护一个窗口。详见「滑动窗口模板」。


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int removeDuplicates(vector<int>& nums) {
    if (nums.empty()) return 0;
    int slow = 0;
    for (int fast = 1; fast < nums.size(); fast++) {
        if (nums[fast] != nums[slow]) {
            slow++;
            nums[slow] = nums[fast];
        }
    }
    return slow + 1;
}
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void moveZeroes(vector<int>& nums) {
    int slow = 0;
    for (int fast = 0; fast < nums.size(); fast++) {
        if (nums[fast] != 0) {
            swap(nums[slow], nums[fast]);
            slow++;
        }
    }
}

类型指针运动方式典型题
快慢指针同方向,速度不同链表环、中点、倒数第k个
左右指针两端向中间两数之和、盛水、三数之和、回文
滑动窗口同方向,维护窗口子串、子数组问题
原地修改慢存快扫去重、移动零

  • 链表问题 → 先想快慢指针
  • 有序数组求和 → 先想左右指针
  • 子数组/子串 → 先想滑动窗口
  • 原地修改数组 → 先想快慢指针(慢存快扫)

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