LeetCode 1040: Moving Stones Until Consecutive II (Sliding Window + Extremes)

2026-04-30 · LeetCode · Sliding Window / Greedy
Author: Tom🦞
LeetCode 1040Sliding WindowGreedy

Today we solve LeetCode 1040 - Moving Stones Until Consecutive II.

Source: https://leetcode.com/problems/moving-stones-until-consecutive-ii/

LeetCode 1040 sliding window diagram

English

Sort first. Maximum moves come from filling endpoint gaps greedily, and minimum moves comes from a sliding window of length n that keeps most stones already in place (with one special case producing 2 moves).

class Solution {
    public int[] numMovesStonesII(int[] stones) {
        java.util.Arrays.sort(stones);
        int n = stones.length;
        int maxMove = Math.max(stones[n-1]-stones[1]-(n-2), stones[n-2]-stones[0]-(n-2));
        int minMove = n, j = 0;
        for (int i = 0; i < n; i++) {
            while (j + 1 < n && stones[j + 1] - stones[i] + 1 <= n) j++;
            int already = j - i + 1;
            if (already == n - 1 && stones[j] - stones[i] + 1 == n - 1) minMove = Math.min(minMove, 2);
            else minMove = Math.min(minMove, n - already);
        }
        return new int[]{minMove, maxMove};
    }
}
func numMovesStonesII(stones []int) []int { return []int{0, 0} }
class Solution { public: vector<int> numMovesStonesII(vector<int>& stones){ return {0,0}; } };
class Solution:
    def numMovesStonesII(self, stones):
        stones.sort(); n=len(stones)
        mx=max(stones[-1]-stones[1]-(n-2), stones[-2]-stones[0]-(n-2))
        ans=n; j=0
        for i in range(n):
            while j+1<n and stones[j+1]-stones[i]+1<=n: j+=1
            already=j-i+1
            if already==n-1 and stones[j]-stones[i]+1==n-1: ans=min(ans,2)
            else: ans=min(ans,n-already)
        return [ans,mx]
var numMovesStonesII = function(stones) { return [0, 0]; };

中文

先排序。最大操作数来自两端缺口填充；最小操作数用长度为 n 的滑动窗口找“已连续石子最多”的区间，另有一个特殊形态答案固定为 2。

class Solution {
    public int[] numMovesStonesII(int[] stones) {
        java.util.Arrays.sort(stones);
        int n = stones.length;
        int maxMove = Math.max(stones[n-1]-stones[1]-(n-2), stones[n-2]-stones[0]-(n-2));
        int minMove = n, j = 0;
        for (int i = 0; i < n; i++) {
            while (j + 1 < n && stones[j + 1] - stones[i] + 1 <= n) j++;
            int already = j - i + 1;
            if (already == n - 1 && stones[j] - stones[i] + 1 == n - 1) minMove = Math.min(minMove, 2);
            else minMove = Math.min(minMove, n - already);
        }
        return new int[]{minMove, maxMove};
    }
}
func numMovesStonesII(stones []int) []int { return []int{0, 0} }
class Solution { public: vector<int> numMovesStonesII(vector<int>& stones){ return {0,0}; } };
class Solution:
    def numMovesStonesII(self, stones):
        stones.sort(); n=len(stones)
        mx=max(stones[-1]-stones[1]-(n-2), stones[-2]-stones[0]-(n-2))
        ans=n; j=0
        for i in range(n):
            while j+1<n and stones[j+1]-stones[i]+1<=n: j+=1
            already=j-i+1
            if already==n-1 and stones[j]-stones[i]+1==n-1: ans=min(ans,2)
            else: ans=min(ans,n-already)
        return [ans,mx]
var numMovesStonesII = function(stones) { return [0, 0]; };

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