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最优算法:贪心法，时间复杂度 O(n) 用一个变量保存你最多可以跳多远• 次优算法:动态规划，时间复杂度 O(n^2) • state: f[i]代表我能否跳到第i个位置• function: f[i] = OR{f[j]} 其中 j < i && j能够跳到i • 解释:什么是 OR 运算?• 比如满足 j < i && j 能够跳到 i 的 j 有 0, 1, 4, 7 • 那么 f[i] = f[0] || f[1] || f[4] || f[7]• initialize: f[0] = true; • answer: f[n-1] int reach = 0; for(int i=0; i<nums.length; i++){ if(reach < i) return false; reach = Math.max(reach, nums[i]+i); } return true;
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