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{
"data": {
"question": {
"questionId": "1000560",
"questionFrontendId": "LCP 72",
"categoryTitle": "Algorithms",
"boundTopicId": 2238419,
"title": "补给马车",
"titleSlug": "hqCnmP",
"content": "English description is not available for the problem. Please switch to Chinese.",
"translatedTitle": "补给马车",
"translatedContent": "远征队即将开启未知的冒险之旅,不过在此之前,将对补给车队进行最后的检查。`supplies[i]` 表示编号为 `i` 的补给马车装载的物资数量。\n考虑到车队过长容易被野兽偷袭他们决定将车队的长度变为原来的一半向下取整计划为\n- 找出车队中 **物资之和最小** 两辆 **相邻** 马车,将它们车辆的物资整合为一辆。若存在多组物资之和相同的马车,则取编号最小的两辆马车进行整合;\n- 重复上述操作直到车队长度符合要求。\n\n请返回车队长度符合要求后物资的分布情况。\n\n**示例 1**\n>输入:`supplies = [7,3,6,1,8]`\n>\n>输出:`[10,15]`\n>\n>解释:\n> 第 1 次合并,符合条件的两辆马车为 6,1合并后的车队为 [7,3,7,8]\n> 第 2 次合并,符合条件的两辆马车为 (7,3) 和 (3,7),取编号最小的 (7,3),合并后的车队为 [10,7,8]\n> 第 3 次合并,符合条件的两辆马车为 7,8合并后的车队为 [10,15]\n>返回 `[10,15]`\n\n**示例 2**\n>输入:`supplies = [1,3,1,5]`\n>\n>输出:`[5,5]`\n\n**解释:**\n- `2 <= supplies.length <= 1000`\n- `1 <= supplies[i] <= 1000`",
"isPaidOnly": false,
"difficulty": "Easy",
"likes": 8,
"dislikes": 0,
"isLiked": null,
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{
"lang": "C++",
"langSlug": "cpp",
"code": "class Solution {\npublic:\n vector<int> supplyWagon(vector<int>& supplies) {\n\n }\n};",
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"lang": "Python",
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"code": "class Solution(object):\n def supplyWagon(self, supplies):\n \"\"\"\n :type supplies: List[int]\n :rtype: List[int]\n \"\"\"",
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{
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"code": "class Solution:\n def supplyWagon(self, supplies: List[int]) -> List[int]:",
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"lang": "C",
"langSlug": "c",
"code": "/**\n * Note: The returned array must be malloced, assume caller calls free().\n */\nint* supplyWagon(int* supplies, int suppliesSize, int* returnSize){\n\n}",
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{
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"code": "public class Solution {\n public int[] SupplyWagon(int[] supplies) {\n\n }\n}",
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{
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"code": "/**\n * @param {number[]} supplies\n * @return {number[]}\n */\nvar supplyWagon = function(supplies) {\n\n};",
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"code": "class Solution {\n\n /**\n * @param Integer[] $supplies\n * @return Integer[]\n */\n function supplyWagon($supplies) {\n\n }\n}",
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"code": "class Solution {\n List<int> supplyWagon(List<int> supplies) {\n\n }\n}",
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