{
   "question_id": 55,
   "question_title": "跳跃游戏",
   "difficulty": "中等",
   "question_content": "<p>给定一个非负整数数组 <code>nums</code> ，你最初位于数组的 <strong>第一个下标</strong> 。</p><p>数组中的每个元素代表你在该位置可以跳跃的最大长度。</p><p>判断你是否能够到达最后一个下标。</p><p> </p><p><strong>示例 1：</strong></p><pre><strong>输入：</strong>nums = [2,3,1,1,4]<strong><br />输出：</strong>true<strong><br />解释：</strong>可以先跳 1 步，从下标 0 到达下标 1, 然后再从下标 1 跳 3 步到达最后一个下标。</pre><p><strong>示例 2：</strong></p><pre><strong>输入：</strong>nums = [3,2,1,0,4]<strong><br />输出：</strong>false<strong><br />解释：</strong>无论怎样，总会到达下标为 3 的位置。但该下标的最大跳跃长度是 0 ， 所以永远不可能到达最后一个下标。</pre><p> </p><p><strong>提示：</strong></p><ul>\t<li><code>1 <= nums.length <= 3 * 10<sup>4</sup></code></li>\t<li><code>0 <= nums[i] <= 10<sup>5</sup></code></li></ul>",
   "topic_link": "https://bbs.csdn.net/topics/600470122",
   "cpp": "#include <stdio.h>\n#include <stdlib.h>\n#include <stdbool.h>\nstatic inline int max(int a, int b)\n{\n\treturn a > b ? a : b;\n}\nstatic bool canJump(int *nums, int numsSize)\n{\n\tint i, pos = 0;\n\tfor (i = 0; i < numsSize - 1; i++)\n\t{\n\t\tif (pos < i || pos >= numsSize - 1)\n\t\t{\n\t\t\tbreak;\n\t\t}\n\t\tpos = max(i + nums[i], pos);\n\t}\n\treturn pos >= numsSize - 1;\n}\nint main(int argc, char **argv)\n{\n\tint i, count = argc - 1;\n\tint *nums = malloc(count * sizeof(int));\n\tfor (i = 0; i < count; i++)\n\t{\n\t\tnums[i] = atoi(argv[i + 1]);\n\t}\n\tprintf(\"%s\\n\", canJump(nums, count) ? \"true\" : \"false\");\n\treturn 0;\n}",
   "java": "class Solution {\n\tpublic boolean canJump(int[] nums) {\n\t\tboolean can = true;\n\t\tif (nums.length < 2) {\n\t\t\treturn can;\n\t\t}\n\t\tint n = nums.length;\n\t\tint stride = 1;\n\t\tfor (int i = n - 2; i >= 0; i--) {\n\t\t\tif (nums[i] < stride) {\n\t\t\t\tstride++;\n\t\t\t\tcan = false;\n\t\t\t} else {\n\t\t\t\tcan = true;\n\t\t\t\tstride = 1;\n\t\t\t}\n\t\t}\n\t\treturn can;\n\t}\n}\n",
   "js": "",
   "python": "class Solution(object):\n\tdef canJump(self, nums):\n\t\t\"\"\"\n\t\t:type nums: List[int]\n\t\t:rtype: bool\n\t\t\"\"\"\n\t\tlength = len(nums)\n\t\tbegin = length - 1\n\t\tfor i in reversed(range(length - 1)):\n\t\t\tif i + nums[i] >= begin:\n\t\t\t\tbegin = i\n\t\treturn not begin\n# %%\ns = Solution()\nprint(s.canJump(nums = [2,3,1,1,4]))",
   "status": 1,
   "keywords": "贪心,数组,动态规划",
   "license": {
      "cpp": "https://github.com/begeekmyfriend/leetcode",
      "python": "https://github.com/qiyuangong/leetcode",
      "java": "https://github.com/zhangyu345293721/leetcode"
   },
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