leetcode-背包子集-416

/**
<p>给你一个 <strong>只包含正整数 </strong>的 <strong>非空 </strong>数组 <code>nums</code> 。请你判断是否可以将这个数组分割成两个子集，使得两个子集的元素和相等。</p>

<p> </p>

<p><strong>示例 1：</strong></p>

<pre>
<strong>输入：</strong>nums = [1,5,11,5]
<strong>输出：</strong>true
<strong>解释：</strong>数组可以分割成 [1, 5, 5] 和 [11] 。</pre>

<p><strong>示例 2：</strong></p>

<pre>
<strong>输入：</strong>nums = [1,2,3,5]
<strong>输出：</strong>false
<strong>解释：</strong>数组不能分割成两个元素和相等的子集。
</pre>

<p> </p>

<p><strong>提示：</strong></p>

<ul>
	<li><code>1 <= nums.length <= 200</code></li>
	<li><code>1 <= nums[i] <= 100</code></li>
</ul>
<div><div>Related Topics</div><div><li>数组</li><li>动态规划</li></div></div><br><div><li>👍 1376</li><li>👎 0</li></div>
*/

//leetcode submit region begin(Prohibit modification and deletion)
class Solution {
    public boolean canPartition(int[] nums) {
        if(nums.length==0){
            return false;
        }

        int total =0;
        for (int i = 0; i < nums.length; i++) {
            total+=nums[i];
        }

        if(total%2!=0){
            return false;
        }

        int amount = total/2;
        int n = nums.length;
        boolean[][] dp = new boolean[n+1][amount+1];
        for (int i = 0; i <=n ; i++) {
            dp[i][0]=true;
        }
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <=amount ; j++) {
                if(j-nums[i-1]>=0){
                    //由于是boolean 所以用或运算
                    dp[i][j]=dp[i-1][j] || dp[i-1][j-nums[i-1]];
                }else{
                    dp[i][j]=dp[i-1][j];
                }
            }
        }
        return dp[n][amount];
    }
}
//leetcode submit region end(Prohibit modification and deletion)