leetcode-dp-509

/**
<p><strong>斐波那契数</strong>&nbsp;（通常用&nbsp;<code>F(n)</code> 表示）形成的序列称为 <strong>斐波那契数列</strong> 。该数列由&nbsp;<code>0</code> 和 <code>1</code> 开始，后面的每一项数字都是前面两项数字的和。也就是：</p>

<pre>
F(0) = 0，F(1)&nbsp;= 1
F(n) = F(n - 1) + F(n - 2)，其中 n &gt; 1
</pre>

<p>给定&nbsp;<code>n</code> ，请计算 <code>F(n)</code> 。</p>

<p>&nbsp;</p>

<p><strong>示例 1：</strong></p>

<pre>
<strong>输入：</strong>n = 2
<strong>输出：</strong>1
<strong>解释：</strong>F(2) = F(1) + F(0) = 1 + 0 = 1
</pre>

<p><strong>示例 2：</strong></p>

<pre>
<strong>输入：</strong>n = 3
<strong>输出：</strong>2
<strong>解释：</strong>F(3) = F(2) + F(1) = 1 + 1 = 2
</pre>

<p><strong>示例 3：</strong></p>

<pre>
<strong>输入：</strong>n = 4
<strong>输出：</strong>3
<strong>解释：</strong>F(4) = F(3) + F(2) = 2 + 1 = 3
</pre>

<p>&nbsp;</p>

<p><strong>提示：</strong></p>

<ul>
	<li><code>0 &lt;= n &lt;= 30</code></li>
</ul>
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*/

//leetcode submit region begin(Prohibit modification and deletion)
class Solution {
    public static int fib(int n) {
        if (n == 0) {
            return 0;
        }

        //基础值
        int[] dp = new int[n + 1];
        dp[0] = 0;
        dp[1] = 1;
        //状态
        for (int i = 2; i <= n; i++) {
            dp[i] = dp[i - 1] + dp[i - 2];
        }
        return dp[n];
    }
}
//leetcode submit region end(Prohibit modification and deletion)