update

leetcode-cn/originData/jump-game-vi.json

@@ -12,7 +12,7 @@
             "translatedContent": "<p>给你一个下标从 <strong>0</strong> 开始的整数数组 <code>nums</code> 和一个整数 <code>k</code> 。</p>\n\n<p>一开始你在下标 <code>0</code> 处。每一步，你最多可以往前跳 <code>k</code> 步，但你不能跳出数组的边界。也就是说，你可以从下标 <code>i</code> 跳到 <code>[i + 1， min(n - 1, i + k)]</code> <strong>包含</strong> 两个端点的任意位置。</p>\n\n<p>你的目标是到达数组最后一个位置（下标为 <code>n - 1</code> ），你的 <strong>得分</strong> 为经过的所有数字之和。</p>\n\n<p>请你返回你能得到的 <strong>最大得分</strong> 。</p>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<b>输入：</b>nums = [<strong>1</strong>,<strong>-1</strong>,-2,<strong>4</strong>,-7,<strong>3</strong>], k = 2\n<b>输出：</b>7\n<b>解释：</b>你可以选择子序列 [1,-1,4,3] （上面加粗的数字），和为 7 。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [<strong>10</strong>,-5,-2,<strong>4</strong>,0,<strong>3</strong>], k = 3\n<b>输出：</b>17\n<b>解释：</b>你可以选择子序列 [10,4,3] （上面加粗数字），和为 17 。\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<b>输入：</b>nums = [1,-5,-20,4,-1,3,-6,-3], k = 2\n<b>输出：</b>0\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li> <code>1 <= nums.length, k <= 10<sup>5</sup></code></li>\n\t<li><code>-10<sup>4</sup> <= nums[i] <= 10<sup>4</sup></code></li>\n</ul>\n",
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             "hints": [
                 "Let dp[i] be \"the maximum score to reach the end starting at index i\". The answer for dp[i] is nums[i] + max{dp[i+j]} for 1 <= j <= k. That gives an O(n*k) solution.",
                 "Instead of checking every j for every i, keep track of the largest dp[i] values in a heap and calculate dp[i] from right to left. When the largest value in the heap is out of bounds of the current index, remove it and keep checking."