DP

字符串编辑
数组区间
类型的dp问题
先找规律
然后列dp方程

Author: h2pl

src/DP/字符串编辑/修改一个字符串为另一个字符串.java

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+package DP.字符串编辑;
+
+/**
+ * Created by 周杰伦 on 2018/4/9.
+ */
+public class 修改一个字符串为另一个字符串 {
+    public int minDistance(String word1, String word2) {
+        if (word1.equals("") && word2.equals(""))return 0;
+        int m = word1.length();
+        int n = word2.length();
+        //把word1的前i个字符变成word2的前j个字符。
+        int [][]dp = new int[m + 1][n + 1];
+        for (int i = 1;i <= m;i ++) {
+            dp[i][0] = i;
+        }
+        for (int j = 1;j <= n;j ++) {
+            dp[0][j] = j;
+        }
+        dp[0][0] = 0;
+        for (int i = 1;i <= m;i ++) {
+            for (int j = 1;j <= n;j ++) {
+                int c1 = dp[i][j - 1] + 1;
+                int c2 = dp[i - 1][j] + 1;
+                int c3 = dp[i - 1][j - 1] + 1;
+                if (word1.charAt(i - 1) == word2.charAt(j - 1)) {
+                    c3 = dp[i - 1][j - 1];
+                }
+                int min = Math.min(Math.min(c1, c2), c3);
+                dp[i][j] = min;
+            }
+        }
+        return dp[m][n];
+    }
+}

src/DP/字符串编辑/字符串编辑.java

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+package DP.字符串编辑;
+
+/**
+ * Created by 周杰伦 on 2018/4/9.
+ */
+public class 字符串编辑 {
+    public static void main(String[] args) {
+        String a = "eat";
+        String b = "sea";
+        System.out.println(minDistance(a, b));
+        System.out.println(LCS(a, b));
+    }
+    public static int minDistance(String word1, String word2) {
+        int len = LCS(word1, word2);
+        return word1.length() + word2.length() - len - len;
+    }
+    public static int LCS(String a, String b) {
+        if (a.equals("") || b.equals(""))return 0;
+        //a的前i个字符与b的前j个字符的公共子序列长度
+        int [][]dp = new int[a.length() + 1][b.length() + 1];
+        for (int i = 0;i < dp.length;i ++) {
+            dp[i][0] = 0;
+        }
+        for (int i = 0;i < dp[0].length;i ++) {
+            dp[0][i] = 0;
+        }
+        for (int i = 1;i <= a.length();i ++) {
+            for (int j = 1;j <= b.length();j ++) {
+                if (a.charAt(i - 1) == b.charAt(j - 1)) {
+                    dp[i][j] = dp[i - 1][j - 1] + 1;
+                }else {
+                    dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
+                }
+            }
+        }
+        return dp[a.length()][b.length()];
+    }
+}

src/DP/数组区间/子数组最大的和.java

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+package DP.数组区间;
+
+/**
+ * Created by 周杰伦 on 2018/4/9.
+ */
+public class 子数组最大的和 {
+    public static void main(String[] args) {
+        int []a = {-2,1,-3,4,-1,2,1,-5,4};
+        System.out.println(maxSubArray(a));
+    }
+    public static int maxSubArray(int[] nums) {
+        int []dp = new int[nums.length + 1];
+        dp[0] = 0;
+        for (int i = 1;i <= nums.length;i ++) {
+            dp[i] = Math.max(dp[i - 1] + nums[i - 1],nums[i - 1]);
+        }
+        int max = dp[1];
+        for (int i = 1;i <= nums.length;i ++) {
+            if (dp[i] > max)max = dp[i];
+        }
+        return max;
+    }
+}

src/DP/数组区间/数组中等差递增子区间的个数.java

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+package DP.数组区间;
+
+/**
+ * Created by 周杰伦 on 2018/4/9.
+ */
+public class 数组中等差递增子区间的个数 {
+    //首先找规律。3个等差数字 dp[3] = 1 .dp[4] = 4个数和后面3个数=2.
+    //dp[5] = 前五个数 2345 345 = 3.可得dp[i] = dp[i - 1] + 1;
+    public static void main(String[] args) {
+
+    }
+    public int numberOfArithmeticSlices(int[] A) {
+        int []dp = new int[A.length];
+        for (int i = 2;i < A.length;i ++) {
+            if (A[i - 1] - A[i - 2] == A[i] - A[i - 1]) {
+                //此处的初值为1，接下来的推导都是正确的。
+                dp[i] = dp[i - 1] + 1;
+            }
+        }
+        int res = 0;
+        //算出总的方法数。妙啊
+        for (int cnt : dp) {
+            res += cnt;
+        }
+        return res;
+    }
+
+// dfs行不通
+// static int count = 0;
+//    public int numberOfArithmeticSlices(int[] A) {
+//        for (int i = 2;i < A.length;i ++) {
+//            dfs(A, i,true, 0);
+//        }
+//        return count;
+//    }
+//    public void dfs(int []A, int cur, boolean flag, int diff) {
+//        if (flag && cur <= A.length) {
+//            count ++;
+//            return;
+//        }
+//        if (A[cur + 1] - A[cur] == A[cur + 2] - A[cur + 1] ) {
+//            flag = true;
+//            diff = A[cur + 2] - A[cur + 1]
+//        }
+//    }
+}

src/DP/数组区间/数组区间和.java

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+package DP.数组区间;
+
+/**
+ * Created by 周杰伦 on 2018/4/9.
+ */
+public class 数组区间和 {
+    class NumArray {
+        private int[] sums;
+
+        // 求区间 i ~ j 的和，可以转换为 sum[j] - sum[i-1]，其中 sum[i] 为 0 ~ i 的和。
+        public NumArray(int[] nums) {
+            sums = new int[nums.length];
+            for (int i = 0; i < nums.length; i++) {
+                sums[i] = i == 0 ? nums[0] : sums[i - 1] + nums[i];
+            }
+        }
+
+        public int sumRange(int i, int j) {
+            return i == 0 ? sums[j] : sums[j] - sums[i - 1];
+        }
+    }
+}