双指针技巧总结

一、双指针技巧——情景1

  通常，我们只需要一个指针进行迭代，即从数组中的第一个元素开始，最后一个元素结束。然而，有时我们会使用两个指针进行迭代。



双指针的典型场景
(1)从两端向中间迭代数组。
(2)一个指针从头部开始，而另一个指针从尾部开始。

1.反转字符串

解法：双指针

c

void swap(char *a, char *b) {
    char t = *a;
    *a = *b, *b = t;
}

void reverseString(char *s, int sSize) {
    for (int left = 0, right = sSize - 1; left < right; ++left, --right) {
        swap(s + left, s + right);
    }
}

c++

class Solution {
public:
    void reverseString(vector<char>& s) {
        int n = s.size();
        for (int left = 0, right = n - 1; left < right; ++left, --right) {
            swap(s[left], s[right]);
        }
    }
};

2.数组拆分

解法：排序

C

int cmp(int *a, int *b) {
    return *a - *b;
}

int arrayPairSum(int *nums, int numsSize) {
    qsort(nums, numsSize, sizeof(int), cmp);
    int ans = 0;
    for (int i = 0; i < numsSize; i += 2) {
        ans += nums[i];
    }
    return ans;
}

解析：
1.qsort函数
qsort是C语言标准库中用于对数组进行快速排序的函数。它的函数原型如下：

void qsort(void *base, size_t nitems, size_t size, int (*compar)(const void *, const void*));

qsort函数的第一个参数是需要排序的数组的指针base，第二个参数是数组的元素个数nitems，第三个参数是每个元素的大小size（以字节为单位），第四个参数是一个指向比较函数的指针compar。

比较函数compar接受两个指向待比较元素的指针，并返回一个整数值，表示它们的大小关系。如果第一个元素小于第二个元素，则返回一个负数；如果它们相等，则返回0；如果第一个元素大于第二个元素，则返回一个正数。这个函数的原型如下：

int (*compar)(const void *, const void*)

C++

class Solution {
public:
    int arrayPairSum(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        int ans = 0;
        for (int i = 0; i < nums.size(); i += 2) {
            ans += nums[i];
        }
        return ans;
    }
};

3.两数之和(输入有序数组）

方法1：二分法查找

c

int* twoSum(int* numbers, int numbersSize, int target, int* returnSize) {
    int* ret = (int*)malloc(sizeof(int) * 2);
    *returnSize = 2;

    for (int i = 0; i < numbersSize; ++i) {
        int low = i + 1, high = numbersSize - 1;
        while (low <= high) {
            int mid = (high - low) / 2 + low;
            if (numbers[mid] == target - numbers[i]) {
                ret[0] = i + 1, ret[1] = mid + 1;
                return ret;
            } else if (numbers[mid] > target - numbers[i]) {
                high = mid - 1;
            } else {
                low = mid + 1;
            }
        }
    }
    ret[0] = -1, ret[1] = -1;
    return ret;
}

解释:
函数 twoSum 接受四个参数：
numbers：一个指向整数数组开头的指针
numbersSize：表示 numbers 数组的大小的整数
target：目标整数
returnSize：一个指向整数的指针，用于返回结果数组的大小

c++

class Solution {
public:
    vector<int> twoSum(vector<int>& numbers, int target) {
        for (int i = 0; i < numbers.size(); ++i) {
            int low = i + 1, high = numbers.size() - 1;
            while (low <= high) {
                int mid = (high - low) / 2 + low;
                if (numbers[mid] == target - numbers[i]) {
                    return {i + 1, mid + 1};
                } else if (numbers[mid] > target - numbers[i]) {
                    high = mid - 1;
                } else {
                    low = mid + 1;
                }
            }
        }
        return {-1, -1};
    }
};

方法2：双指针

c

int* twoSum(int* numbers, int numbersSize, int target, int* returnSize) {
    int* ret = (int*)malloc(sizeof(int) * 2);
    *returnSize = 2;

    int low = 0, high = numbersSize - 1;
    while (low < high) {
        int sum = numbers[low] + numbers[high];
        if (sum == target) {
            ret[0] = low + 1, ret[1] = high + 1;
            return ret;
        } else if (sum < target) {
            ++low;
        } else {
            --high;
        }
    }
    ret[0] = -1, ret[1] = -1;
    return ret;
}

c++

class Solution {
public:
    vector<int> twoSum(vector<int>& numbers, int target) {
        int low = 0, high = numbers.size() - 1;
        while (low < high) {
            int sum = numbers[low] + numbers[high];
            if (sum == target) {
                return {low + 1, high + 1};
            } else if (sum < target) {
                ++low;
            } else {
                --high;
            }
        }
        return {-1, -1};
    }
};

二、双指针技巧——情景2

1.移除元素

方法1：双指针

c

int removeElement(int* nums, int numsSize, int val) {
    int left = 0;
    for (int right = 0; right < numsSize; right++) {
        if (nums[right] != val) {
            nums[left] = nums[right];
            left++;
        }
    }
    return left;
}

c++

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

方法2：双指针优化

c

int removeElement(int* nums, int numsSize, int val) {
    int left = 0, right = numsSize;
    while (left < right) {
        if (nums[left] == val) {
            nums[left] = nums[right - 1];
            right--;
        } else {
            left++;
        }
    }
    return left;
}

c++

class Solution {
public:
    int removeElement(vector<int>& nums, int val) {
        int left = 0, right = nums.size();
        while (left < right) {
            if (nums[left] == val) {
                nums[left] = nums[right - 1];
                right--;
            } else {
                left++;
            }
        }
        return left;
    }
};

2.最大连续1的个数

方法1：一次遍历

c

int findMaxConsecutiveOnes(int* nums, int numsSize) {
    int maxCount = 0, count = 0;
    for (int i = 0; i < numsSize; i++) {
        if (nums[i] == 1) {
            count++;
        } else {
            maxCount = fmax(maxCount, count);
            count = 0;
        }
    }
    maxCount = fmax(maxCount, count);
    return maxCount;
}

c++

class Solution {
public:
    int findMaxConsecutiveOnes(vector<int>& nums) {
        int maxCount = 0, count = 0;
        int n = nums.size();
        for (int i = 0; i < n; i++) {
            if (nums[i] == 1) {
                count++;
            } else {
                maxCount = max(maxCount, count);
                count = 0;
            }
        }
        maxCount = max(maxCount, count);
        return maxCount;
    }
};

3.长度最小的子数组

方法1：暴力法

c

int minSubArrayLen(int s, int* nums, int numsSize) {
    if (numsSize == 0) {
        return 0;
    }
    int ans = INT_MAX;
    for (int i = 0; i < numsSize; i++) {
        int sum = 0;
        for (int j = i; j < numsSize; j++) {
            sum += nums[j];
            if (sum >= s) {
                ans = fmin(ans, j - i + 1);
                break;
            }
        }
    }
    return ans == INT_MAX ? 0 : ans;
}

C++

class Solution {
public:
    int minSubArrayLen(int s, vector<int>& nums) {
        int n = nums.size();
        if (n == 0) {
            return 0;
        }
        int ans = INT_MAX;
        for (int i = 0; i < n; i++) {
            int sum = 0;
            for (int j = i; j < n; j++) {
                sum += nums[j];
                if (sum >= s) {
                    ans = min(ans, j - i + 1);
                    break;
                }
            }
        }
        return ans == INT_MAX ? 0 : ans;
    }
};

方法2：前缀和+二分查找

c

int lower_bound(int *a, int l, int r, int q) {
    if (a[r] < q) return -1;
    while (l < r) {
        int mid = (l + r) >> 1;
        if (a[mid] >= q) {
            r = mid;
        } else {
            l = mid + 1;
        }
    }
    return l;
}
int minSubArrayLen(int s, int *nums, int numsSize) {
    if (numsSize == 0) {
        return 0;
    }
    int ans = INT_MAX;
    int *sums = (int *)malloc(sizeof(int) * (numsSize + 1));
    // 为了方便计算，令 size = n + 1
    // sums[0] = 0 意味着前 0 个元素的前缀和为 0
    // sums[1] = A[0] 前 1 个元素的前缀和为 A[0]
    // 以此类推
    for (int i = 1; i <= numsSize; i++) {
        sums[i] = sums[i - 1] + nums[i - 1];
    }
    for (int i = 1; i <= numsSize; i++) {
        int target = s + sums[i - 1];
        int bound = lower_bound(sums, 1, numsSize, target);
        if (bound != -1) {
            ans = fmin(ans, bound - (i - 1));
        }
    }
    return ans == INT_MAX ? 0 : ans;
}

C++

class Solution {
public:
    int minSubArrayLen(int s, vector<int>& nums) {
        int n = nums.size();
        if (n == 0) {
            return 0;
        }
        int ans = INT_MAX;
        vector<int> sums(n + 1, 0); 
        // 为了方便计算，令 size = n + 1 
        // sums[0] = 0 意味着前 0 个元素的前缀和为 0
        // sums[1] = A[0] 前 1 个元素的前缀和为 A[0]
        // 以此类推
        for (int i = 1; i <= n; i++) {
            sums[i] = sums[i - 1] + nums[i - 1];
        }
        for (int i = 1; i <= n; i++) {
            int target = s + sums[i - 1];
            auto bound = lower_bound(sums.begin(), sums.end(), target);
            if (bound != sums.end()) {
                ans = min(ans, static_cast<int>((bound - sums.begin()) - (i - 1)));
            }
        }
        return ans == INT_MAX ? 0 : ans;
    }
};

方法3：滑动窗口

c

int minSubArrayLen(int s, int *nums, int numsSize) {
    if (numsSize == 0) {
        return 0;
    }
    int ans = INT_MAX;
    int start = 0, end = 0;
    int sum = 0;
    while (end < numsSize) {
        sum += nums[end];
        while (sum >= s) {
            ans = fmin(ans, end - start + 1);
            sum -= nums[start];
            start++;
        }
        end++;
    }
    return ans == INT_MAX ? 0 : ans;
}

C++

class Solution {
public:
    int minSubArrayLen(int s, vector<int>& nums) {
        int n = nums.size();
        if (n == 0) {
            return 0;
        }
        int ans = INT_MAX;
        int start = 0, end = 0;
        int sum = 0;
        while (end < n) {
            sum += nums[end];
            while (sum >= s) {
                ans = min(ans, end - start + 1);
                sum -= nums[start];
                start++;
            }
            end++;
        }
        return ans == INT_MAX ? 0 : ans;
    }
};