DP总结

• 最长回文子序列

  int lpsDp(char * str,int n){
    int dp[n][n], tmp;
    memset(dp,0,sizeof(dp));
    for(int i=0; i<n; i++) dp[i][i] = 1;
    // i 表示 当前长度为 i+1的 子序列
    for(int i=1; i<n; i++){
        tmp = 0;
        //考虑所有连续的长度为i+1的子串. 该串为 str[j, j+i]
        for(int j=0; j+i<n; j++){
            //如果首尾相同
            if(str[j] == str[j+i]){
                tmp = dp[j+1][j+i-1] + 2;
            }else{
                tmp = max(dp[j+1][j+i],dp[j][j+i-1]);
            }
            dp[j][j+i] = tmp;
        }
    }
    //返回串 str[0][n-1] 的结果
    return dp[0][n-1];
  }
  

• 最长递增子序列

  public class Solution {
    public int lengthOfLIS(int[] nums) {
        if(nums==null || nums.length == 0) return 0;
        int len = nums.length;
        int[] dp = new int[len];
        int[] ends = new int[len];
        ends[0] = nums[0];
        dp[0] = 1;
        int right = 0;
        int l=0;
        int r=0;
        int m=0;      
        for(int i=0;i<len;i++){
            l = 0;
            r = right;
            while(l<=r){
                m = l + (r-l)/2;
                if(nums[i]>ends[m]){
                    l = m + 1;
                }else{
                    r = m - 1;
                }
            }
            right = Math.max(right,l);
            ends[l] = nums[i];
            dp[i] = l + 1;
        }
        int res = 0;
        for(int i=0;i<len;i++){
            res=  Math.max(res,dp[i]);
        }
        return res;
    }
  }
  

• Distinct Subsequences

  public class Solution {
    public int numDistinct(String s, String t) {
        int len1 = s.length();
        int len2 = t.length();
        int[][] dp = new int[len2+1][len1+1];
        for(int j=0;j<=len1;j++){dp[0][j] = 1;}
        for(int i=0;i<len2;i++){ 
            for(int j=0;j<len1;j++){
                if(t.charAt(i) == s.charAt(j)){
                    dp[i+1][j+1] = dp[i][j]+dp[i+1][j];
                }else{
                    dp[i+1][j+1] = dp[i+1][j];
                }
            }
        }
        return dp[len2][len1];
    }
  }
  

• Decode Ways

  public class Solution {
    public int numDecodings(String s) {
        if(s==null || s.length()==0) return 0;
        char[] ca = s.toCharArray();
        int n = ca.length;
        if(n==1) return (ca[0]=='0')?0:1;
        int[] dp = new int[n];
        dp[0] = (ca[0]=='0')?0:1;
        dp[1] = ((ca[0]!='0' && ca[1]!='0')?1:0) + (((ca[0]!='0')&&(ca[0]-'0')*10+(ca[1]-'0')<=26)?1:0);
        if(n>2){
            for(int i=2;i<n;i++){
                dp[i] = 0;
                if(ca[i]!='0'){
                    dp[i]+=dp[i-1];
                }
                if(ca[i-1]!='0' && ((ca[i-1]-'0')*10+(ca[i]-'0')<=26) ){
                    dp[i]+=dp[i-2];
                }
            }
        }
        return dp[n-1];
    }
  }
  

• Edit Distance

  public class Solution {
    public int minDistance(String word1, String word2) {
        int m = word1.length();
        int n = word2.length();
  
        int[][] dp = new int[m + 1][n + 1];
        for(int i=0;i<=n;i++){dp[0][i] = i;}
        for(int i=0;i<=m;i++){dp[i][0] = i;}
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                if(word1.charAt(i) == word2.charAt(j)){
                    dp[i+1][j+1] = dp[i][j];
                }else{
                    dp[i+1][j+1] = Math.min(Math.min(dp[i+1][j],dp[i][j+1]),dp[i][j])+1;
                }
            }
        }
        return dp[m][n];
    }
  }
  

• Minimum Path Sum

  public class Solution {
    public int minPathSum(int[][] grid) {
        int m = grid.length;
        int n = grid[0].length;
        int[][] dp = new int[m][n];
        dp[0][0] = grid[0][0];
        for(int i=1;i<m;i++){
            dp[i][0] = dp[i-1][0]+grid[i][0];
        }
        for(int j=1;j<n;j++){
            dp[0][j] = dp[0][j-1]+grid[0][j];
        }
  
        for(int i=1;i<m;i++){
            for(int j=1;j<n;j++){
                dp[i][j] = Math.min(dp[i-1][j],dp[i][j-1])+grid[i][j];
            }
        }
        return dp[m-1][n-1];
    }
  }
  

• Maximum Subarray

  public class Solution {
    public int maxSubArray(int[] nums) {
        int currMax=nums[0],max=nums[0];
        for(int i=1;i<nums.length;i++){
            currMax=Math.max(currMax+nums[i],nums[i]);
            max=Math.max(max,currMax);
        }
        return max;
    }
  }
  

• Unique Paths II

  public class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        int width = obstacleGrid[0].length;
        int[] dp = new int[width];
        dp[0] = 1;
        for (int[] row : obstacleGrid) {
            for (int j = 0; j < width; j++) {
                if (row[j] == 1)
                    dp[j] = 0;
                else if (j > 0)
                    dp[j] += dp[j - 1];
            }
        }
        return dp[width - 1];
    }
  }
  

• Wildcard Matching

  boolean[] match = new boolean[m+1];
  match[0] = true;
  for (int i = 0; i < m; i++) {
    match[i+1] = false;
  }
  for (int i = 0; i < n; i++) {
    if (p.charAt(i) == '*') {
        for (int j = 0; j < m; j++) {
            match[j+1] = match[j] || match[j+1]; 
        }
    } else {
        for (int j = m-1; j >= 0; j--) {
            match[j+1] = (p.charAt(i) == '?' || p.charAt(i) == s.charAt(j)) && match[j];
        }
        match[0] = false;
    }
  }
  return match[m];
  

• Regular Expression Matching

  public class Solution {
    public boolean isMatch(String s, String p) {
        boolean[] match = new boolean[s.length()+1];
        Arrays.fill(match,false);
        match[s.length()] = true;
        for(int i=p.length()-1;i>=0;i--){
            if(p.charAt(i) == '*'){
                for(int j=s.length()-1;j>=0;j--){
                    match[j] = match[j] ||match[j+1]&&(p.charAt(i-1)=='.'||s.charAt(j)==p.charAt(i-1)); 
                }
                i--;
            }else{
                for(int j=0;j<s.length();j++){
                    match[j] = match[j+1]&&(p.charAt(i)=='.'||p.charAt(i)==s.charAt(j));
                }
                match[s.length()] = false;
            }
        }
        return match[0];
    }
  }
  

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