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37 lines
1.3 KiB
HTML
37 lines
1.3 KiB
HTML
<p>Given a <code>triangle</code> array, return <em>the minimum path sum from top to bottom</em>.</p>
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<p>For each step, you may move to an adjacent number of the row below. More formally, if you are on index <code>i</code> on the current row, you may move to either index <code>i</code> or index <code>i + 1</code> on the next row.</p>
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<p> </p>
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<p><strong class="example">Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> triangle = [[2],[3,4],[6,5,7],[4,1,8,3]]
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<strong>Output:</strong> 11
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<strong>Explanation:</strong> The triangle looks like:
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<u>2</u>
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<u>3</u> 4
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6 <u>5</u> 7
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4 <u>1</u> 8 3
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The minimum path sum from top to bottom is 2 + 3 + 5 + 1 = 11 (underlined above).
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</pre>
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<p><strong class="example">Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> triangle = [[-10]]
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<strong>Output:</strong> -10
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</pre>
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<p> </p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 <= triangle.length <= 200</code></li>
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<li><code>triangle[0].length == 1</code></li>
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<li><code>triangle[i].length == triangle[i - 1].length + 1</code></li>
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<li><code>-10<sup>4</sup> <= triangle[i][j] <= 10<sup>4</sup></code></li>
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</ul>
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<p> </p>
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<strong>Follow up:</strong> Could you do this using only <code>O(n)</code> extra space, where <code>n</code> is the total number of rows in the triangle? |