我们定义bool dp[i+1]为s(0~i)是否可以被分解，这样就能很容易得到递推式并求解
相对上一题目而言，这道题目的结果变成了一个全路径的搜索问题；不仅要判断s是否可以分解，还要给出所有分解的方案.这算是DP问题的一种很典型的形式。如果我们用vector<vector<string> > dp[i+1]来存储解决方案，这样会导致空间的溢出O(n^3).所以，改进一下，这里我们并不存储全路径，而是存储部分路径。vector<int> dp[i+1], 能够匹配到dp[i+1]的上一个下标,这样空间复杂度是O(n^2).
class Solution {
public:
vector<string> wordBreak(string s, unordered_set<string> &dict) {
VS result;
int n=s.size();
if(n==0){
result.push_back("");
return result;
}
VI test;//dp[i]from 0 to n: the last part of path
VVI dp( n+1, test);
int i, t;
for (i = 0; i < n; ++i){
for (t = 0; t <i+1; ++t){
if(t==0 || dp[t].size()!=0){//s[t-1] is ok!
if(dict.find(s.substr(t, i+1 - t)) != dict.end()){
dp[i+1].push_back(t);//forcus
}
}
}
}
string path;
if(dp[n].size()!=0)
result = bfs(n, dp, path, s);
return result;
}
VS bfs(int index, VVI const &dp, string path, string const &s){
VS result;
if(dp[index].size()==0){
result.push_back(path.substr(1, path.length()));
return result;
}
int n= dp[index].size();
int i;
for (i = 0; i < n; ++i){
VS pre;
pre = bfs(dp[index][i], dp, " "+ s.substr(dp[index][i], index - dp[index][i])+path, s ) ;
int size_pre = pre.size();
int j;
for(j=0; j<size_pre; j++){
result.push_back(pre[j]);
}
}
return result;
}
};
Eg 3:Palindrome Partitioning 动态规划 N/A
class Solution {
public:
vector<vector<string>> partition(string s) {
int n = s.length();
vector<vector<string> > res;
vector<string> each;
if(n < 2) {
each.push_back(s);
res.push_back(each);
return res;
}
/* get state[i][j] */
vector<bool> done(n, false);
vector<vector<bool> > state(n, done);
get_palindron_state(s, state);
vector<string> pathed;
dfs(s, state, n-1, pathed, res);
return res;
}
void get_palindron_state(string s, vector<vector<bool> > &state){
int n = state.size();
for (int i = n-1; i >=0 ; --i){
for (int j = i; j < n; ++j){
if(i==j) state[i][j]=true;
else{
if(s[i] == s[j]){
if(i+1==j) state[i][j]=true;
else state[i][j]= state[i+1][j-1];
}
}
}
}
}
void dfs(const string &s,const vector<vector<bool> > &state, int end, vector<string> pathed, VVS &res ){
for (int i = end; i >= 0; --i){
if(state[i][end]){
pathed.insert(pathed.begin(), s.substr(i, end-i+1));
dfs(s,state, i-1, pathed, res);
pathed.erase(pathed.begin());
}
}
if(end == -1){
res.push_back(pathed);
}
}
}; Eg4:Palindrome Partitioning II
dp[i] = min{dp[j+1] + 1| s(i, j)是回文}
Eg5:Triangle
太典型了，不讲。
Eg6:,Distinct Subsequences
分析：这道题目，可能初看之下，不能很好的想到使用动态规划求解—— 这是一个双序列类型的题目，熟练了以后就比较好往DP思考了。现在来看一下分析过程。
还是那句话：DP问题的难点不在于构造而在于识别。如果不能一眼看出DP问题的特征，那么就老老实实按照：问题是否可分；分解后子问题是否重叠；重叠的子问题是否满足最优子结构来判定。
class Solution {
public:
int numDistinct(string S, string T) {
int i, j;
int m= S.length(), n=T.length();
int dp[m][n];
if(m==0 || n==0 ) return 0;
fill(dp[0], dp[0]+n, 0);
if(S[0]==T[0]) dp[0][0]=1;
for (i = 1; i < m; ++i){
dp[i][0]=dp[i-1][0];
if(S[i]==T[0]) dp[i][0]++;
}
for (i = 1; i < m; ++i){
for(j=1; j<n; j++){
dp[i][j] = dp[i-1][j];
if(S[i]== T[j]) dp[i][j] += dp[i-1][j-1];
}
}
return dp[m-1][n-1];
}
}; class Solution {
public:
int numDecodings(string s) {
int n=s.length();
if(n==0 ) return 0;
int i;
int dp[n+1];
fill(dp, dp+n+1, 0);
dp[0]=1;
if(isdigit(s[0] ) && s[0]!='0') dp[1]= 1;
else dp[1]=0;
if(n==1) return dp[1];
for (i = 1; i < n; ++i){
if(isdigit(s[i]) && s[i]!='0') dp[i+1] += dp[i];
if(s.substr(i-1, 2) >= string("10") && s.substr(i-1, 2) <= string("26")) dp[i+1] += dp[i-1];
}
return dp[n];
}
}; Eg8:Scramble String
这是这个问题的递归解法，也是最直观的解法；首先，子问题可分是肯定的;另外，通过对子问题的观察，我们发现的确有子问题重叠现象，现在就剩下如何定义DP了。
class Solution {
public:
bool isScramble(string s1, string s2) {
int n1=s1.length();
int n2=s2.length();
if(n1!=n2)
return false;
bool dp[n1][n1][n1+1];
memset(dp, false, sizeof(dp));
/* dp[i][j][k] start of s1, j:start of s2, k len of them */
int i, j, k, l;
for (i = 0; i < n1; ++i){
for (j = 0; j < n1; ++j){
dp[i][j][1] = (s1[i]==s2[j]);
}
}
for(k=2; k<=n1;k++){
for(i=0; i+k<=n1; i++){
for (j = 0; j+k <= n1; ++j){
for(l=1; l<k; l++){
if( (dp[i][j][l] && dp[i+l][j+l][k-l]) || (dp[i][j+k-l][l] && dp[i+l][j][k-l])){
dp[i][j][k] = true;
break;
}
}
}
}
}
return dp[0][0][n1];
}
}; Eg9:Maximal Rectangle
这道题，DP是其中一个比较小的环节，用户数据预处理，dp[i][j]第i行j列上的元素往上延伸的1的个数。剩下的，就是另外一道题目的变形：最大直方图。
class Solution {
public:
int maximalRectangle(vector<vector<char> > &matrix) {
int row_len = matrix.size();
if(row_len == 0) return 0;
int col_len = matrix[0].size();
if(col_len ==0 ) return 0;
vector<int> height(col_len, 0);
vector<int> lastheight(col_len, 0);
int res=0;
for (int i = 0; i < row_len; ++i){
for( int j=0; j<col_len ; j++){
if(matrix[i][j] == '1'){
height[j]=lastheight[j]+1;
}
else
height[j]=0;
}
res = max(res, largest(height));
lastheight = height;
}
return res;
}
int largest(vector<int> height) {
height.push_back(0);// be cautious
stack<int> stk;
int i = 0;
int maxArea = 0;
while(i < height.size()){
if(stk.empty() || height[stk.top()] <= height[i]){
stk.push(i++);
}else {
int t = stk.top();
stk.pop();
maxArea = max(maxArea, height[t] * (stk.empty() ? i : i - stk.top() -1));
}
}
return maxArea;
}
}; Eg9:Edit Distance
class Solution {
public:
int minDistance(string word1, string word2) {
int len1 = word1.length(), len2 = word2.length();
int dp[len1+1][len2+1];
for (int i = 0; i < len1+1; ++i){
dp[i][0]=i;
}
for (int i = 0; i < len2+1; ++i){
dp[0][i]=i;
}
int i, j;
for (i = 0; i < len1; ++i){
for (j = 0; j < len2; ++j){
if(word1[i] == word2[j]) dp[i+1][j+1] = dp[i][j];
else {
dp[i+1][j+1] = min(dp[i][j], min(dp[i+1][j], dp[i][j+1])) +1 ;
}
}
}
return dp[len1][len2];
}
}; 动态规划第五讲——leetcode上的题目动态规划汇总（上）
原文地址：http://blog.csdn.net/trochiluses/article/details/38017259
