揹包-力扣-java

class Solution {
    public boolean canPartition(int[] nums) {
        int sum=0;
        for(int num:nums)
            sum+=num;
        if(sum%2==1) return false;
        sum=sum/2;
        int len=nums.length;
        boolean[][] dp = new boolean[len+1][sum+1];
        for(int i=0;i<=len;i++) dp[i][0]=true;
        for(int i=1;i<=len;i++){
            for(int j=1;j<=sum;j++){
                if(j<nums[i-1]){   //// 揹包容量不足，不能裝入第 i 個物品
                    dp[i][j]=dp[i-1][j];
                }
                else{
                    dp[i][j] = dp[i-1][j] || dp[i-1][j-nums[i-1]];
                }
            }
        }  
        return dp[len][sum];  
    }
}
//https://leetcode-cn.com/problems/partition-equal-subset-sum/solution/0-1-bei-bao-wen-ti-bian-ti-zhi-zi-ji-fen-ge-by-lab/

 

class Solution {
    public boolean wordBreak(String s, List<String> wordDict) {
        //可以類比於揹包問題
        int n=s.length();
        //pointer[i]表示s中以i-1結尾的字符串是否可以被拆分
        boolean[] pointer=new boolean[n+1];
        pointer[0]=true;
        for(int i=1;i<=n;i++){
            for(int j=0;j<i;j++){
                if(pointer[j]&&wordDict.contains(s.substring(j,i))){
                    pointer[i]=true;
                    break;
                }
            }
        }
        return pointer[n];
    }
}