一、基础算法精讲：双指针

1、相向双指针 1

1.1 两数之和 II - 输入有序数组

Leetcode 167

注意，这里可以使用相向双指针的原因是因为这里的数组是非递减的。

class Solution:
    def twoSum(self, numbers: List[int], target: int) -> List[int]:
        l = 0
        r = len(numbers) - 1
        while True:
            s = numbers[l] + numbers[r]
            if s == target:
                return [l + 1, r + 1]
            if s > target:
                r -= 1
            else:
                l += 1

class Solution {
public:
    vector<int> twoSum(vector<int>& numbers, int target) {
        int l = 0, r = numbers.size() - 1;
        while (true) {
            int s = numbers[l] + numbers[r];
            if (s == target) return {l + 1, r + 1};
            s > target ? r -- : l ++ ;
        }
    }
};

• 时间复杂度：$O(n)$
• 空间复杂度：$O(1)$

1.2 三数之和

Leetcode 15

class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]:
        nums.sort()
        ans = []
        n = len(nums)
        for i in range(n - 2):
            x = nums[i]
            if i > 0 and x == nums[i - 1]:  # 跳过重复数字
                continue
            if x + nums[i + 1] + nums[i + 2] > 0:  # 优化一
                break
            if x + nums[-2] + nums[-1] < 0:  # 优化二
                continue
            j = i + 1
            k = n - 1
            while j < k:
                s = x + nums[j] + nums[k]
                if s > 0:
                    k -= 1
                elif s < 0:
                    j += 1
                else:
                    ans.append([x, nums[j], nums[k]])
                    j += 1
                    while j < k and nums[j] == nums[j - 1]:  # 跳过重复数字
                        j += 1
                    k -= 1
                    while k > j and nums[k] == nums[k + 1]:  # 跳过重复数字
                        k -= 1
        return ans

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        vector<vector<int>> ans;
        int n = nums.size();
        for (int i = 0; i < n - 2; i ++ ) {
            int x = nums[i];
            if (i && x == nums[i - 1]) continue;  // 跳过重复数字
            if (x + nums[i + 1] + nums[i + 2] > 0) break;  // 优化一
            if (x + nums[n - 1] + nums[n - 2] < 0) continue;  // 优化二
            int j = i + 1, k = n - 1;
            while (j < k) {
                int s = x + nums[j] + nums[k];
                if (s > 0) k -- ;
                else if (s < 0) j ++ ;
                else {
                    ans.push_back({x, nums[j], nums[k]});
                    for ( ++ j; j < k && nums[j] == nums[j - 1]; j ++ );  // 跳过重复元素
                    for ( -- k; k > j && nums[k] == nums[k + 1]; k -- );  // 跳过重复元素
                }
            }
        }
        return ans;
    }
}; 

• 时间复杂度：$O(n^2)$
• 空间复杂度：$O(1)$

1.3 最接近的三数之和

Leetcode 16

注意这里的三个优化：

class Solution:
    def threeSumClosest(self, nums: List[int], target: int) -> int:
        nums.sort()
        n = len(nums)
        minDiff = inf
        ans = 0

        for i in range(n - 2):
            x = nums[i]

            # 优化三
            if i and x == nums[i - 1]: 
                continue

            # 优化一
            s = x + nums[i + 1] + nums[i + 2]
            if s > target:
                if s - target < minDiff:
                    minDiff = s - target
                    ans = s
                break
            
            # 优化二
            s = x + nums[-1] + nums[-2]
            if s < target:
                if target - s < minDiff:
                    minDiff = target - s
                    ans = s
                continue

            j, k = i + 1, n - 1
            while j < k:
                s = x + nums[j] + nums[k]

                if s == target:
                    return s

                if s > target:      
                    if s - target < minDiff:
                        minDiff = s - target
                        ans = s      
                    k -= 1 
                else:
                    if target - s < minDiff:
                        minDiff = target - s
                        ans = s
                    j += 1

        return ans

class Solution {
public:
    int threeSumClosest(vector<int>& nums, int target) {
        int res;
        int min_diff = 1e9;
        int n = nums.size();
        sort(nums.begin(), nums.end());
        for (int i = 0; i < n - 2; i ++ ) {
            int x = nums[i];

            // 优化一
            if (i && x == nums[i - 1]) continue;

            // 优化二
            int s = x + nums[i + 1] + nums[i + 2];
            if (s > target) {
                if (s - target < min_diff) 
                    min_diff = s - target, res = s;
                break;
            }
            
            // 优化三
            s = x + nums[n - 2] + nums[n - 1];
            if (s < target) {
                if (target - s < min_diff)
                    min_diff = target - s, res = s;
                continue;
            }

            int j = i + 1, k = n - 1;
            while (j < k) {
                s = x + nums[j] + nums[k];
                if (s == target) return s;
                if (s > target) {
                    if (s - target < min_diff)
                        min_diff = s - target, res = s;
                        k -- ;
                } else {
                    if (target - s < min_diff)
                        min_diff = target - s, res = s;
                    j ++ ;
                }
            }
        }
        return res;
    }
};

• 时间复杂度：$O(n^2)$
• 空间复杂度：$O(1)$

1.4 四数之和

Leetcode 18

class Solution:
    def fourSum(self, nums: List[int], target: int) -> List[List[int]]:
        nums.sort()
        n = len(nums)
        ans = []
        for a in range(n - 3):
            x = nums[a]
            if a and x == nums[a - 1]: 
                continue
            if x + nums[a + 1] + nums[a + 2] + nums[a + 3] > target: 
                break
            if x + nums[-1] + nums[-2] + nums[-3] < target: 
                continue;
            for b in range(a + 1, n - 2):
                y = nums[b]
                if b > a + 1 and y == nums[b - 1]: 
                    continue
                if x + y + nums[b + 1] + nums[b + 2] > target: 
                    break
                if x + y + nums[-1] + nums[-2] < target: 
                    continue
                
                c, d = b + 1, n - 1
                while c < d:
                    s = x + y + nums[c] + nums[d]
                    if s > target: 
                        d -= 1
                    elif s < target: 
                        c += 1
                    else: 
                        ans.append([x, y, nums[c], nums[d]])
                        c += 1
                        while c < d and nums[c] == nums[c - 1]: 
                            c += 1
                        d -= 1
                        while c < d and nums[d] == nums[d + 1]: 
                            d -= 1
        return ans

class Solution {
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target) {
        vector<vector<int>> ans;
        int n = nums.size();
        sort(nums.begin(), nums.end());
        for (int a = 0; a < n - 3; a ++ ) {
            long long x = nums[a];
            if (a && x == nums[a - 1]) continue;
            if (x + nums[a + 1] + nums[a + 2] + nums[a + 3] > target) break;
            if (x + nums[n - 1] + nums[n - 2] + nums[n - 3] < target) continue;
            for (int b = a + 1; b < n - 2; b ++ ) {
                long long y = nums[b];
                if (b > a + 1 && y == nums[b - 1]) continue;
                if (x + y + nums[b + 1] + nums[b + 2] > target) break;
                if (x + y + nums[n - 1] + nums[n - 2] < target) continue;
                int c = b + 1, d = n - 1;
                while (c < d) {
                    long long s = x + y + nums[c] + nums[d];
                    if (s < target) c += 1;
                    else if (s > target) d -= 1;
                    else {
                        ans.push_back({(int)x, (int)y, nums[c], nums[d]});
                        for (c ++ ; c < d && nums[c] == nums[c - 1]; c ++ );
                        for (d -- ; c < d && nums[d] == nums[d + 1]; d -- );
                    }
                }  
            }
        }
        return ans;
    }
};

• 时间复杂度：$O(n^3)$
• 空间复杂度：$O(1)$

1.5 统计和小于目标的下标对数目

Leetcode 2824

class Solution:
    def countPairs(self, nums: List[int], target: int) -> int:
        nums.sort()
        n = len(nums)
        l, r = 0, n - 1
        ans = 0
        while l < r:
            if nums[l] + nums[r] < target:
                ans += r - l
                l += 1
            else:
                r -= 1
        return ans

class Solution {
public:
    int countPairs(vector<int>& nums, int target) {
        int n = nums.size();
        sort(nums.begin(), nums.end());
        int l = 0, r = n - 1, ans = 0;
        while (l < r) 
            if (nums[l] + nums[r] < target)
                ans += r - l, l += 1;
            else r -= 1;
        return ans;
    }
};

• 时间复杂度：$O(n)$
• 空间复杂度：$O(1)$

1.6 有效三角形的个数

Leetcode 611

class Solution:
    def triangleNumber(self, nums: List[int]) -> int:
        nums.sort()
        n = len(nums)
        ans = 0
        for k in range(2, n):
            c = nums[k]
            l, r = 0, k - 1
            while l < r:
                if nums[l] + nums[r] > c:
                    ans += r - l
                    r -= 1
                else:
                    l += 1
        return ans

class Solution {
public:
    int triangleNumber(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        int n = nums.size();
        int ans = 0;
        for (int k = 2; k < n; k ++ ) {
            int c = nums[k];
            int l = 0, r = k - 1;
            while (l < r) 
                if (nums[l] + nums[r] > c)
                    ans += r - l, r -= 1;
                else l += 1;
        }
        return ans;
    }
};

• 时间复杂度：$O(n^2)$
• 空间复杂度：$O(1)$

2、相向双指针 2

2.1 盛最多水的容器

Leetcode 11

class Solution:
    def maxArea(self, height: List[int]) -> int:
        ans = l = 0
        r = len(height) - 1
        while l < r:
            area = (r - l) * min(height[l], height[r])
            ans = max(ans, area)
            if height[l] < height[r]:
                l += 1
            else:
                r -= 1
        return ans

class Solution {
public:
    int maxArea(vector<int>& height) {
        int ans = 0, l = 0, r = height.size() - 1;
        while (l < r) {
            int area = (r - l) * min(height[l], height[r]);
            ans = max(area, ans);
            if (height[l] < height[r]) l += 1;
            else r -= 1;
        }
        return ans;
    }
};

• 时间复杂度：$O(n)$
• 空间复杂度：$O(1)$

2.2 接雨水

Leetcode 42

解法一：前后缀分解

class Solution:
    def trap(self, height: List[int]) -> int:
        n = len(height)
        pre_max = [0] * n
        pre_max[0] = height[0]
        for i in range(1, n):
            pre_max[i] = max(height[i], pre_max[i - 1])
        
        suf_max = [0] * n
        suf_max[-1] = height[-1]
        for i in range(n - 2, -1, -1):
            suf_max[i] = max(height[i], suf_max[i + 1])

        ans = 0
        for h, pre, suf in zip(height, pre_max, suf_max):
            ans += min(pre, suf) - h
        return ans

class Solution {
public:
    int trap(vector<int>& height) {
        int n = height.size();
        vector<int> pre_max(n);
        pre_max[0] = height[0];
        for (int i = 1; i < n; i ++ )
            pre_max[i] = max(height[i], pre_max[i - 1]);
        
        vector<int> suf_max(n);
        suf_max[n - 1] = height[n - 1];
        for (int i = n - 2; i >= 0; i -- )
            suf_max[i] = max(height[i], suf_max[i + 1]);
        
        int ans = 0;
        for (int i = 0; i < n; i ++ ) 
            ans += min(pre_max[i], suf_max[i]) - height[i];
        return ans;
    }
};

• 时间复杂度：$O(n)$
• 空间复杂度：$O(n)$

解法二：双指针

class Solution:
    def trap(self, height: List[int]) -> int:
        ans = l = pre_max = suf_max = 0
        r = len(height) - 1
        while l < r:
            pre_max = max(pre_max, height[l])
            suf_max = max(suf_max, height[r])
            if pre_max < suf_max:
                ans += pre_max - height[l]
                l += 1
            else:
                ans += suf_max - height[r]
                r -= 1
        return ans

class Solution {
public:
    int trap(vector<int>& height) {
        int l = 0, ans = 0, pre_max = 0, suf_max = 0, r = height.size() - 1;
        while (l < r) {
            pre_max = max(pre_max, height[l]);
            suf_max = max(suf_max, height[r]);
            if (pre_max < suf_max)
                ans += pre_max - height[l], l += 1;
            else ans += suf_max - height[r], r -= 1;
        }
        return ans;
    }
};

• 时间复杂度：$O(n)$
• 空间复杂度：$O(1)$

解法三：单调栈

注意：一个凹槽由三个位置决定【最左边（$height[left]$），中间（$botto{m}_h$），最右边（$h$）】

class Solution:
    def trap(self, height: List[int]) -> int:
        ans = 0
        st = []
        for i, h in enumerate(height):
            while st and h >= height[st[-1]]:  # 表示可以形成一个凹槽
                bottom_h = height[st.pop()]    # 弹出栈顶元素并赋值给bottom_h
                if not st:                     # 没有左边界
                    break
                left = st[-1]
                dh = min(height[left], h) - bottom_h  # 高度差
                ans += dh * (i - left - 1)
            st.append(i)
        return ans

class Solution {
public:
    int trap(vector<int>& height) {
        int ans = 0;
        stack<int> st;
        for (int i = 0; i < height.size(); i ++ ) {
            while (!st.empty() && height[i] >= height[st.top()]) {
                int bottom_h = height[st.top()];
                st.pop();
                if (st.empty()) break;
                int left = st.top();
                int dh = min(height[left], height[i]) - bottom_h;
                ans += dh * (i - left - 1);
            }
            st.push(i);
        }
        return ans;
    }
};

• 时间复杂度：$O(n)$
• 空间复杂度：$O(min(U,n))$，其中 $U=max(height)-min(height)+1$

3、同向双指针：滑动窗口（区间大小可变）

3.1 长度最小的子数组

Leetcode 209

这里能够使用双指针是因为 $while$ 条件中是满足单调性的

class Solution:
    def minSubArrayLen(self, target: int, nums: List[int]) -> int:
        n = len(nums)
        ans = inf
        s = left = 0
        for right, x in enumerate(nums):
            s += x
            while s >= target:
                ans = min(ans, right - left + 1)
                s -= nums[left]
                left += 1
        return ans if ans <= n else 0

class Solution {
public:
    int minSubArrayLen(int target, vector<int>& nums) {
        int n = nums.size(), ans = n + 1, s = 0, left = 0;
        for (int right = 0; right < n; right ++ ) {
            s += nums[right];
            while (s >= target) {
                ans = min(ans, right - left + 1);
                s -= nums[left ++ ];
            }
        }
        return ans <= n ? ans : 0;
    }
};

• 时间复杂度：$O(n)$
• 空间复杂度：$O(1)$

3.2 乘积小于 K 的子数组

Leetocde 713

class Solution:
    def numSubarrayProductLessThanK(self, nums: List[int], k: int) -> int:
        if k <= 1: return 0
        ans = l = 0
        prod = 1
        for r, x in enumerate(nums):
            prod *= x
            while prod >= k:
                prod /= nums[l]
                l += 1
            ans += r - l + 1
        return ans

class Solution {
public:
    int numSubarrayProductLessThanK(vector<int>& nums, int k) {
        if (k <= 1) return 0;
        int n = nums.size(), ans = 0, prod = 1, left = 0;
        for (int right = 0; right < n; right ++ ) {
            prod *= nums[right];
            while (prod >= k) prod /= nums[left ++ ];
            ans += right - left + 1;
        }
        return ans;
    }
};

• 时间复杂度：$O(n)$
• 空间复杂度：$O(1)$

3.3 无重复字符的最长字串

Leetcode 3

class Solution:
    def lengthOfLongestSubstring(self, s: str) -> int:
        ans = l = 0
        w = set()
        for r, c in enumerate(s):
            while c in w:
                w.remove(s[l])
                l += 1
            w.add(c)
            ans = max(ans, r - l + 1)
        return ans

python 另一个版本，效率比上面一个略微差一点：

class Solution:
    def lengthOfLongestSubstring(self, s: str) -> int:
        ans = 0
        cnt = Counter()
        left = 0
        for right, c in enumerate(s):
            cnt[c] += 1
            while cnt[c] > 1:
                cnt[s[left]] -= 1
                left += 1
            ans = max(ans, right - left + 1)
        return ans

class Solution {
public:
    int lengthOfLongestSubstring(string s) {
        int n = s.size(), ans = 0, l = 0;
        unordered_set<char> w;
        for (int r = 0; r < n; r ++ ) {
            char c = s[r];
            while (w.count(c)) w.erase(s[l ++ ]);
            w.insert(c);
            ans = max(ans, r - l + 1);
        }
        return ans;
    }
};

• 时间复杂度：$O(n)$
• 空间复杂度：$O(128)\approx O(1)$

3.4 最大连续1的个数 III

Leetcode 1004

class Solution:
    def longestOnes(self, nums: List[int], k: int) -> int:
        ans = left = cnt0 = 0  # cnt0 用于统计窗口内部0的个数
        for right, x in enumerate(nums):
            cnt0 += 1 - x
            while cnt0 > k:
                cnt0 -= 1 - nums[left]
                left += 1
            ans = max(ans, right - left + 1)
        return ans

class Solution {
public:
    int longestOnes(vector<int>& nums, int k) {
        int n = nums.size(), l = 0, ans = 0, cnt0 = 0;
        for (int r = 0; r < n; r ++ ) {
            cnt0 += 1 - nums[r];
            while (cnt0 > k) 
                cnt0 -= 1 - nums[l ++ ];
            ans = max(ans, r - l + 1);
        }
        return ans;
    }
};

• 时间复杂度：$O(n)$
• 空间复杂度：$O(1)$

3.5 替换子串得到平衡字符串

Leetcode 1234

如果在 待替换子串（即滑动窗口） 之外的任意字符的出现次数超过 $m=\frac{n}{4}$，那么无论怎么替换，都无法使这个字符的出现次数等于 $m$。

反过来说，如果在待替换子串之外的任意字符的出现次数都不超过 $m$，那么可以通过替换，使 $s$ 为平衡字符串，即每个字符的出现次数均为 $m$。

class Solution:
    def balancedString(self, s: str) -> int:
        cnt, m = Counter(s), len(s) // 4

        # 用于判断可迭代对象中的所有元素是否都为真
        if all(cnt[x] == m for x in "QWER"): 
            return 0
            
        ans, left = inf, 0
        for right, c in enumerate(s):
            cnt[c] -= 1
            while all(cnt[x] <= m for x in "QWER"):
                ans = min(ans, right - left + 1)
                cnt[s[left]] += 1
                left += 1
                
        return ans

class Solution {
public:
    int balancedString(string s) {
        int n = s.length(), m = n / 4, cnt['X']{};
        for (char c: s) cnt[c] ++ ;
        if (cnt['Q'] == m && cnt['W'] == m && cnt['E'] == m && cnt['R'] == m)
            return 0;
        int ans = n, l = 0;
        for (int r = 0; r < n; r ++ ) {
            cnt[s[r]] -- ;
            while (cnt['Q'] <= m && cnt['W'] <= m && cnt['E'] <= m && cnt['R'] <= m) {
                ans = min(ans, r - l + 1);
                cnt[s[l ++ ]] ++ ;
            }
        }
        return ans;
    }
};

• 时间复杂度：$O(CN)$，其中 $C=4$，$N$ 为字符串 $s$ 的长度
• 空间复杂度：$O(C)$

3.6 将 x 减到 0 的最小操作数

Leetcode 1658

逆向思维

将原问题转换为求解数组中最长的子数组，使得子数组的元素和为 $s-x$，其中 $s$ 表示整个数组的和。

class Solution:
    def minOperations(self, nums: List[int], x: int) -> int:
        target = sum(nums) - x
        if target < 0: return -1
        ans = -1  # 存储长度
        left = s = 0  # s 存储窗口内的和
        for right, x in enumerate(nums):
            s += x
            while s > target:
                s -= nums[left]
                left += 1
            if s == target:
                ans = max(ans, right - left + 1)
        return -1 if ans < 0 else len(nums) - ans

class Solution {
public:
    int minOperations(vector<int>& nums, int x) {
        int target = accumulate(nums.begin(), nums.end(), 0) - x;
        if (target < 0) return -1;
        int ans = -1, l = 0, sum = 0, n = nums.size();
        for (int r = 0; r < n; r ++ ) {
            sum += nums[r];
            while (sum > target) sum -= nums[l ++ ];
            if (sum == target) ans = max(ans, r - l + 1);
        }        
        return ans < 0 ? -1 : n - ans;
    }
};

• 时间复杂度：$O(n)$
• 空间复杂度：$O(1)$