数组相关题目

没什么技巧。直接暴力解

Author: h2pl

src/数据结构/数组与矩阵/对角元素相等的矩阵.java

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+package 数据结构.数组与矩阵;
+
+/**
+ * Created by 周杰伦 on 2018/4/17.
+ */
+public class 对角元素相等的矩阵 {
+    public static void main(String[] args) {
+        int [][] matrix = {{1,2,3,4},{5,1,2,3},{9,5,1,2}};
+        System.out.println(isToeplitzMatrix(matrix));
+    }
+    //按照每个斜行来判断，一旦有不同的值直接返回false
+    public static boolean isToeplitzMatrix(int[][] matrix) {
+        int m = matrix.length;
+        int n = matrix[0].length;
+        int len = m > n ? n : m;
+        for (int i = 1;i < len;i ++) {
+             if (matrix[i - 1][i - 1] != matrix[i][i]) {
+                 return false;
+             }
+        }
+        for (int i = 1;i < m - 1;i ++) {
+            for (int j = 0;j < n - 1;j ++) {
+                if (matrix[i][j] != matrix[i + 1][j + 1]) {
+                    return false;
+                }
+            }
+        }
+        for (int i = 1;i < n - 1;i ++) {
+            for (int j = 0;j < m - 1;j ++) {
+                if (matrix[j][i] != matrix[j + 1][i + 1]) {
+                    return false;
+                }
+            }
+        }
+        return true;
+    }
+}

src/数据结构/数组与矩阵/找出数组中最长的连续1.java

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+package 数据结构.数组与矩阵;
+
+/**
+ * Created by 周杰伦 on 2018/4/17.
+ */
+public class 找出数组中最长的连续1 {
+    // on
+    public int findMaxConsecutiveOnes(int[] nums) {
+        int max = 0;
+        int count = 0;
+        for (int i = 0;i < nums.length;i ++) {
+            if (nums[i] == 1) {
+                count ++;
+                if (count > max) {
+                    max = count;
+                }
+            }else {
+                count = 0;
+            }
+        }
+        return max;
+    }
+}

src/数据结构/数组与矩阵/数组的度.java

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+package 数据结构.数组与矩阵;
+
+import java.util.HashMap;
+import java.util.Set;
+
+/**
+ * Created by 周杰伦 on 2018/4/17.
+ */
+public class 数组的度 {
+    public static void main(String[] args) {
+        int []a = {1,2,2,3,1,4,2};
+        System.out.println(findShortestSubArray(a));
+    }
+    //暴力解法：
+    //先找到出现次数最多的数。
+    //可能有多个，对每个数求需要序列的长度。取最小值
+    public static int findShortestSubArray(int[] nums) {
+        int max = 0;
+        int cnt = 0;
+        HashMap<Integer, Integer> map = new HashMap<>();
+        for (int i = 0;i < nums.length;i ++) {
+            if (map.containsKey(nums[i])) {
+                map.put(nums[i], map.get(nums[i]) + 1);
+            }else {
+                map.put(nums[i], 1);
+            }
+        }
+        Set<Integer> set = map.keySet();
+        for (int i : set) {
+            max = Math.max(map.get(i), max);
+        }
+        System.out.println("max:" + max);
+        int min = Integer.MAX_VALUE;
+        for (int x : set) {
+            cnt = 0;
+            int k = -1;
+            for (int i = 0; i < nums.length; i++) {
+                if (x == nums[i]) {
+                    cnt ++;
+                    if (k == -1) {
+                        k = i;
+                    }
+                }
+                if (cnt == max) {
+                    min = Math.min(i - k + 1, min);
+                    break;
+                }
+            }
+        }
+        return min;
+    }
+}

src/数据结构/数组与矩阵/数组相邻差值的个数.java

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+package 数据结构.数组与矩阵;
+
+import java.util.Arrays;
+
+/**
+ * Created by 周杰伦 on 2018/4/17.
+ */
+public class 数组相邻差值的个数 {
+    //这题不会 注意一下
+    //让前 k+1 个元素构建出 k 个不相同的差值，序列为：1 k+1 2 k 3 k-1 ... k/2 k/2+1.
+    public int[] constructArray(int n, int k) {
+        int[] ret = new int[n];
+        ret[0] = 1;
+        for (int i = 1, interval = k; i <= k; i++, interval--) {
+            ret[i] = i % 2 == 1 ? ret[i - 1] + interval : ret[i - 1] - interval;
+        }
+        for (int i = k + 1; i < n; i++) {
+            ret[i] = i + 1;
+        }
+        return ret;
+    }
+//    public int[] constructArray(int n, int k) {
+//        int []num = new int[n];
+//        int []diff = new int[n];
+//        Arrays.fill(diff, -1);
+//        int x = 1;
+//        for (int j = 1;j <= n;j ++) {
+//            num[0] = j;
+//            int cntk = 0;
+//            for (int i = 2; i <= n; i++) {
+//                if (diff[Math.abs(num[x] - num[x - 1])] == -1 && j != i && cntk < k) {
+//                    num[x ++] = i;
+//                    diff[Math.abs(num[x] - num[x - 1])] = 1;
+//                }
+//                if (k == cntk) {
+//                    num[x ++] = i;
+//                    if (x == n) {
+//                        return num;
+//                    }
+//                }
+//            }
+//        }
+//        return num;
+//    }
+}

src/数据结构/数组与矩阵/调整矩阵.java

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+package 数据结构.数组与矩阵;
+
+/**
+ * Created by 周杰伦 on 2018/4/17.
+ */
+public class 调整矩阵 {
+    //先转存到一维数组，再重新放到新数组。
+    //如果行列乘积不相等，直接返回原数组。
+    public int[][] matrixReshape(int[][] nums, int r, int c) {
+        int m = nums.length,n = nums[0].length;
+        int [][] arr = new int[r][c];
+        int []num = new int[m * n];
+        if (m * n == r * c) {
+            int k = 0;
+            for (int i = 0 ;i < m;i ++) {
+                for (int j = 0;j < n;j ++) {
+                    num[k ++] = nums[i][j];
+                }
+            }
+            k = 0;
+            for (int i = 0;i < r;i ++) {
+                for (int j = 0;j < c;j ++) {
+                    arr[i][j] = num[k ++];
+                }
+            }
+            return arr;
+        }
+        return nums;
+    }
+}