[面试算法] 动态规划
LCS（最长公共子序列）
int LCS(string s1, string s2){
int len1 = s1.size(), len2 = s2.size();
vector> lcs(len1 + 1, vector(len2 + 1));
for (int i = 0; i <= len1; i++)
lcs[i][0] = 0;
for (int i = 0; i <= len2; i++){
lcs[0][i] = 0;
}
for (int i = 1; i <= len1; i++){
for (int j = 1; j <= len2; j++){
if (s1[i-1] == s2[j-1]){
lcs[i][j] = lcs[i - 1][j - 1] + 1;
}
else{
lcs[i][j] = max(lcs[i - 1][j], lcs[i][j - 1]);
}
}
}
return lcs[len1][len2];
} LCS（最长公共子串）
string LCS(string s1, string s2){
int len1 = s1.size(), len2 = s2.size();
vector> lcs(len1 + 1, vector(len2 + 1, 0));
for (int i = 0; i <= len1; i++)
lcs[i][0] = 0;
for (int i = 0; i <= len2; i++){
lcs[0][i] = 0;
}
int last = 0, len = 0;
for (int i = 1; i <= len1; i++){
for (int j = 1; j <= len2; j++){
if (s1[i - 1] == s2[j - 1]){ //注意矩阵和字符串索引的对应
lcs[i][j] = lcs[i - 1][j - 1] + 1;
}
else{
lcs[i][j] = 0;
}
if (lcs[i][j] > len){
len = lcs[i][j];
last = i;
}
}
}
return s1.substr(last - len, len);
} LIS（最长递增子序列）
int lengthOfLIS(vector& nums) {
int maxlen = 0, len = nums.size();
vector dp(len, 1);
for (int i = 0; i < len; i++) {
for (int j = 0; j < i; j++) {
if (nums[i] > nums[j])
dp[i] = max(dp[i], dp[j] + 1);
}
maxlen = max(maxlen, dp[i]);
}
return maxlen;
} 字符串编辑距离（字符串相似度）
int minDistance(string word1, string word2) {
int len1 = word1.size(), len2 = word2.size();
vector> dist(len1 + 1, vector(len2 + 1, 0));
for (int i = 1; i <= len1; i++){
dist[i][0] = i;
}
for (int i = 1; i <= len2; i++){
dist[0][i] = i;
}
int f;
for (int i = 1; i <= len1; i++){
for (int j = 1; j <= len2; j++){
if (word1[i - 1] == word2[j - 1])
f = 0;
else
f = 1;
dist[i][j] = min(dist[i - 1][j] + 1, dist[i][j - 1] + 1);
dist[i][j] = min(dist[i][j], dist[i - 1][j - 1] + f);
}
}
return dist[len1][len2];
} 交替字符串：
bool isInterleave(string s1, string s2, string s3) {
int len1 = s1.size(), len2 = s2.size();
if (len1 + len2 != s3.size())
return false;
vector> dp(len1 + 1, vector(len2 + 1, false));
dp[0][0] = true;
for (int i = 1; i <= len1; i++){
dp[i][0] = dp[i - 1][0] && (s1[i - 1] == s3[i - 1]);
}
for (int j = 1; j <= len2; j++){
dp[0][j] = dp[0][j - 1] && (s2[j - 1] == s3[j - 1]);
}
for (int i = 1; i <= len1; i++){
for (int j = 1; j <= len2; j++){
dp[i][j] = (dp[i][j - 1] && s2[j - 1] == s3[i + j - 1]) || (dp[i - 1][j] && s1[i - 1] == s3[i + j - 1]);
}
}
return dp[len1][len2];
} 子序列数目（Distinct Subsequences）
int numDistinct(string s, string t) {
int len1 = s.size(), len2 = t.size();
vector> dist(len1 + 1, vector(len2 + 1, 0));
for (int i = 0; i <= len1; i++){
dist[i][0] = 1;
}
for (int i = 1; i <= len1; i++){
for (int j = 1; j <= len2; j++){
dist[i][j] = dist[i - 1][j];
if (s[i - 1] == t[j - 1]){
dist[i][j] += dist[i - 1][j - 1];
}
}
}
return dist[len1][len2];
}