leetcode-416-dp
/**
<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)
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