From 961aa4cea9b6691b6dedaa9dedffc264aef0f33f Mon Sep 17 00:00:00 2001 From: whai Date: Fri, 8 Dec 2023 21:22:50 +0800 Subject: [PATCH] =?UTF-8?q?LeetCode997=20=E5=8F=8C=E6=8C=87=E9=92=88?= =?UTF-8?q?=EF=BC=81?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit - 找到0的左右,分别从左右插入到新数组中 - 从左到右比较最大的数,从大到小插入新的数组中 --- .../whaifree/leetCode/easy/LeetCode977.java | 152 ++++++++++++++++++ 1 file changed, 152 insertions(+) create mode 100644 src/main/java/cn/whaifree/leetCode/easy/LeetCode977.java diff --git a/src/main/java/cn/whaifree/leetCode/easy/LeetCode977.java b/src/main/java/cn/whaifree/leetCode/easy/LeetCode977.java new file mode 100644 index 0000000..38a8e18 --- /dev/null +++ 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]); + } + } + +}