diff --git a/src/main/java/cn/whaifree/leetCode/easy/LeetCode977.java b/src/main/java/cn/whaifree/leetCode/easy/LeetCode977.java
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+++ b/src/main/java/cn/whaifree/leetCode/easy/LeetCode977.java
@@ -0,0 +1,152 @@
+package cn.whaifree.leetCode.easy;
+
+import org.junit.Test;
+
+/**
+ * 977. 有序数组的平方
+ *
+ * 给你一个按 非递减顺序 排序的整数数组 nums，返回 每个数字的平方 组成的新数组，要求也按 非递减顺序 排序。
+ *
+ * 示例 1：
+ *
+ * 输入：nums = [-4,-1,0,3,10]
+ * 输出：[0,1,9,16,100]
+ * 解释：平方后，数组变为 [16,1,0,9,100]
+ * 排序后，数组变为 [0,1,9,16,100]
+ *
+ * 示例 2：
+ *
+ * 输入：nums = [-7,-3,2,3,11]
+ * 输出：[4,9,9,49,121]
+ *
+ * 提示：
+ *
+ * 1 <= nums.length <= 104
+ * -104 <= nums[i] <= 104
+ * nums 已按 非递减顺序 排序
+ *
+ *
+ * 进阶：
+ *
+ * 请你设计时间复杂度为 O(n) 的算法解决本问题
+ *
+ * 更多挑战
+ * 88. 合并两个有序数组
+ * 360. 有序转化数组
+ */
+public class LeetCode977 {
+
+    /**
+     * 找到第一个左边小于0，右边大于等于0的点
+     * 左右两个指针
+     * 平方后必然是，【递减 0 递增】
+     * [-3,-1,0,2,5]
+     * [9,1,0,4,25]
+     *
+     * 让左边的循环插入到右边合适的位置
+     * @param nums
+     * @return
+     */
+    public int[] sortedSquares(int[] nums) {
+
+        int middle = nums.length;
+        for (int i = 0; i < nums.length; i++) {
+            if (nums[i] >= 0) {
+                middle = i;
+                break;
+            }
+        }
+        // 此时就有两个数组了
+        // [0,middle-1] 为递减的
+        // [middle, length-1] 为递增的
+        int[] ints = new int[nums.length];
+        int leftIndex = middle -1;
+        int rightIndex = middle;
+        int intIndex = 0;
+        while (leftIndex >= 0 && rightIndex < nums.length) {
+            if (nums[leftIndex] * nums[leftIndex] < nums[rightIndex] * nums[rightIndex]) {
+                ints[intIndex++] = nums[leftIndex] * nums[leftIndex];
+                leftIndex--;
+            } else {
+                ints[intIndex++] = nums[rightIndex] * nums[rightIndex];
+                rightIndex++;
+            }
+        }
+
+        while (leftIndex >= 0) {
+            ints[intIndex++] = nums[leftIndex] * nums[leftIndex];
+            leftIndex--;
+        }
+        while (rightIndex <= nums.length-1) {
+            ints[intIndex++] = nums[rightIndex] * nums[rightIndex];
+            rightIndex++;
+        }
+
+        return ints;
+    }
+
+
+    /**
+     * 不考虑0的双指针
+     * @param nums
+     * @return
+     */
+    public int[] sortedSquares1(int[] nums) {
+        int[] ints = new int[nums.length];
+
+        int left = 0;
+        int right = nums.length - 1;
+        int index = 0;
+        while (left <= right) {
+            if (nums[left] * nums[left] < nums[right] * nums[right]) {
+                ints[index++] = nums[right] * nums[right];
+                right--;
+            } else {
+                ints[index++] = nums[left] * nums[left];
+                left++;
+            }
+        }
+
+        // 逆序
+        int value = 0;
+        for (int i = 0; i < ints.length / 2; i++) {
+            value = ints[i];
+            ints[i] = ints[ints.length - i - 1];
+            ints[ints.length - 1 - i] = value;
+        }
+
+        return ints;
+    }
+
+    public int[] sortedSquares2(int[] nums) {
+        int[] ints = new int[nums.length];
+
+        int left = 0;
+        int right = nums.length - 1;
+        // 存入新数组的索引，从大往小存
+        int index = nums.length - 1;
+        while (left <= right) {
+            // 负数 + 正数 > 0  则 |负数| < |正数|
+            if (nums[left] + nums[right] > 0) {
+                ints[index--] = nums[right] * nums[right];
+                right--;
+            } else {
+                ints[index--] = nums[left] * nums[left];
+                left++;
+            }
+        }
+
+
+        return ints;
+    }
+
+
+    @Test
+    public void test() {
+        int[] ints = sortedSquares2(new int[]{-3,-1,4,10});
+        for (int i = 0; i < ints.length; i++) {
+            System.out.println(ints[i]);
+        }
+    }
+
+}