{"id":5118,"date":"2019-04-28T11:44:27","date_gmt":"2019-04-28T18:44:27","guid":{"rendered":"https:\/\/zxi.mytechroad.com\/blog\/?p=5118"},"modified":"2019-04-28T11:51:22","modified_gmt":"2019-04-28T18:51:22","slug":"leetcode-45-jump-game-ii","status":"publish","type":"post","link":"https:\/\/zxi.mytechroad.com\/blog\/greedy\/leetcode-45-jump-game-ii\/","title":{"rendered":"\u82b1\u82b1\u9171 LeetCode 45. Jump Game II"},"content":{"rendered":"\n

Given an array of non-negative integers, you are initially positioned at the first index of the array.<\/p>\n\n\n\n

Each element in the array represents your maximum jump length at that position.<\/p>\n\n\n\n

Your goal is to reach the last index in the minimum number of jumps.<\/p>\n\n\n\n

Example:<\/strong><\/p>\n\n\n\n

Input:<\/strong> [2,3,1,1,4]\nOutput:<\/strong> 2\nExplanation:<\/strong> The minimum number of jumps to reach the last index is 2.\n    Jump 1 step from index 0 to 1, then 3 steps to the last index.<\/pre>\n\n\n\n
Note:<\/strong><\/p>\n\n\n\n
You can assume that you can always reach the last index.<\/p>\n\n\n\n
Solution: Greedy<\/strong><\/h2>\n\n\n\n
Jump as far as possible but lazily. <\/p>\n\n\n\n
\n[2, 3, 1, 1, 4]\ni    nums[i]   steps   near   far\n-      -         0       0     0\n0      2         0       0     2\n1      3         1       2     4\n2      1         1       2     4\n3      1         2       4     4\n4      4         2       4     8\n<\/pre>\n\n\n\n
Time complexity: O(n)
Space complexity: O(1)<\/p>\n\n\n\n
\n
C++<\/h2>\n
\n\n
\n\/\/ Author: Huahua, running time: 12 ms \/ 10.3 MB\nclass Solution {\npublic:\n  int jump(vector& nums) {\n    int steps = 0;\n    int near = 0;\n    int far = 0;\n    for (int i = 0; i < nums.size(); ++i) {\n      if (i > near) {\n        ++steps;\n        near = far;\n      }\n      far = max(far, i + nums[i]);      \n    }\n    return steps;\n  }\n};\n<\/pre>\n<\/div><\/div>\n\n\n\n
<\/p>\n","protected":false},"excerpt":{"rendered":"
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