algorithm001 · GeekUniversity · May 13, 2019 · May 12, 2019
diff --git a/Week_04/id_102/leetcode_242_102.cpp b/Week_04/id_102/leetcode_242_102.cpp
@@ -0,0 +1,17 @@
+class Solution {
+public:
+    int majorityElement(vector<int>& nums) {
+        /* 建立hash表　key-value */
+        unordered_map<int, int> map;
+        int ret = 0;
+        int len = nums.size();
+        for (int i = 0; i < len; i++) {
+            map[nums[i]]++;
+            if (map[nums[i]]>len/2) {
+                ret = nums[i];
+                break;
+            }
+        }
+        return ret;
+    }
+};
diff --git a/Week_04/id_102/leetcode_72_102.cpp b/Week_04/id_102/leetcode_72_102.cpp
@@ -0,0 +1,27 @@
+class Solution {
+public:
+    int minDistance(string word1, string word2) {
+        int row = word1.size();
+        int column = word2.size();
+
+        int f[row+1][column+1];
+        for (int i = 0; i <= row; i++) {
+            f[i][0] = i;
+        }
+
+        for (int j = 0; j <=column; j++) {
+            f[0][j] = j;
+        }
+
+        for (int i = 1; i <= row; i++) {
+            for (int j = 1; j <= column; j++) {
+                if (word1[i-1] == word2[j-1]) {
+                    f[i][j] = f[i-1][j-1];
+                } else {
+                    f[i][j] = 1 + min(f[i-1][j-1], min(f[i][j-1], f[i-1][j]));
+                }
+            }
+        }
+        return f[row][column];
+    }
+};
diff --git a/Week_04/id_102/leetcode_746_102.cpp b/Week_04/id_102/leetcode_746_102.cpp
@@ -0,0 +1,14 @@
+class Solution {
+public:
+    int minCostClimbingStairs(vector<int>& cost) {
+        vector<int> dp(cost.size(), 0);
+        dp[0] = cost[0];
+        dp[1] = cost[1];
+
+        for (int i = 2; i < cost.size(); i++) {
+            dp[i] = min(dp[i-1], dp[i-2]) + cost[i];
+        }
+
+        return min(dp[dp.size()-1], dp[dp.size()-2]);
+    }
+};
diff --git a/Week_04/id_102/leetcode_784_012.cpp b/Week_04/id_102/leetcode_784_012.cpp
@@ -0,0 +1,53 @@
+/*
+给定一个字符串S，通过将字符串S中的每个字母转变大小写，我们可以获得一个新的字符串。返回所有可能得到的字符串集合。
+
+示例:
+输入: S = "a1b2"
+输出: ["a1b2", "a1B2", "A1b2", "A1B2"]
+
+输入: S = "3z4"
+输出: ["3z4", "3Z4"]
+
+输入: S = "12345"
+输出: ["12345"]
+*/
+
+class Solution {
+public:
+    vector<string> letterCasePermutation(string S) {
+        if(S.empty()) {
+            return {}; 
+        }
+
+        vector<string> vec;
+        Permutation(0, vec, S);
+        return vec;
+    }
+
+private:
+    void Permutation(int idx, vector<string>& ans, string S) {
+        if (idx == S.size()) {
+            ans.push_back(S);
+            return;
+        }
+
+        if (S[idx] >= '0' && S[idx] <= '9') {
+            Permutation(idx + 1, ans, S);
+            return;
+        } 
+
+        if (S[idx] >= 'A' && S[idx] <= 'Z') {
+            S[idx] = tolower(S[idx]);
+        } 
+        Permutation(idx + 1, ans, S);
+
+        if (S[idx] >= 'a' && S[idx] <= 'z') {
+            S[idx] = toupper(S[idx]);
+        }
+        Permutation(idx + 1, ans, S);
+
+        return;
+    }
+};
+
+