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新增java二維"整數"數組的解法、優化一維數組的解法(剪枝)
1. 新增java二維"整數"數組的解法,原本是用boolean數組,有一點繞,故附上二維"整數"數組。 2. 優化一維數組的解法(剪枝),在每一次內層循環完後,立刻檢查dp[target] == target。實測時間複雜度: 26ms -> 20ms
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@ -220,6 +220,10 @@ class Solution {
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//物品 i 的重量是 nums[i],其价值也是 nums[i]
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dp[j] = Math.max(dp[j], dp[j - nums[i]] + nums[i]);
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}
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//剪枝一下,每一次完成內層的for-loop,立即檢查是否dp[target] == target,優化時間複雜度(26ms -> 20ms)
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if(dp[target] == target)
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return true;
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}
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return dp[target] == target;
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}
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@ -294,6 +298,61 @@ false true false false false true true false false false false false
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false true false false false true true false false false false true
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false true false false false true true false false false true true
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```
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二維數組整數版本
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```Java
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class Solution {
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public boolean canPartition(int[] nums) {
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//using 2-D DP array.
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int len = nums.length;
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//check edge cases;
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if(len == 0)
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return false;
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int sum = 0;
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for (int num : nums)
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sum += num;
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//we only deal with even numbers. If sum is odd, return false;
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if(sum % 2 == 1)
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return false;
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int target = sum / 2;
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int[][] dp = new int[nums.length][target + 1];
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// for(int j = 0; j <= target; j++){
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// if(j < nums[0])
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// dp[0][j] = 0;
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// else
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// dp[0][j] = nums[0];
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// }
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//initialize dp array
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for(int j = nums[0]; j <= target; j++){
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dp[0][j] = nums[0];
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}
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for(int i = 1; i < len; i++){
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for(int j = 0; j <= target; j++){
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if (j < nums[i])
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dp[i][j] = dp[i - 1][j];
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else
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dp[i][j] = Math.max(dp[i - 1][j], dp[i - 1][j - nums[i]] + nums[i]);
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}
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}
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//print out DP array
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// for(int x : dp){
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// System.out.print(x + ",");
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// }
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// System.out.print(" "+i+" row"+"\n");
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return dp[len - 1][target] == target;
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}
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}
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//dp数组的打印结果 for test case 1.
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0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
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0, 1, 1, 1, 1, 5, 6, 6, 6, 6, 6, 6,
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0, 1, 1, 1, 1, 5, 6, 6, 6, 6, 6, 11,
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0, 1, 1, 1, 1, 5, 6, 6, 6, 6, 10, 11,
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```
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### Python:
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```python
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