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前言

轮转数组分为左移和右移

一、力扣189. 轮转数组

轮转数组分为左移和右移

class Solution {public void rotate(int[] nums, int k) {if(k == 0){return;}int len = nums.length-(k%nums.length);convert(nums,0,len-1);convert(nums,len,nums.length-1);convert(nums,0,nums.length-1);}public void convert(int[] nums, int left, int right){for(int i = left, j = right; i < j; i ++, j --){int temp = nums[i];nums[i] = nums[j];nums[j] = temp;}}
}

二、力扣724. 寻找数组的中心下标

class Solution {public int pivotIndex(int[] nums) {int sum = 0;for(int a : nums){sum += a;}int leftSum = 0;for(int i = 0; i < nums.length; i ++){if(leftSum*2+nums[i] == sum){return i;}else{leftSum += nums[i];}}return -1;}
}

三、力扣34. 在排序数组中查找元素的第一个和最后一个位置

class Solution {public int[] searchRange(int[] nums, int target) {int[] res = new int[]{-1,-1};if(nums.length == 0){return res;}boolean flag = true;for(int i = 0; i < nums.length; i ++){if(nums[i] == target){if(flag){res[0] = i;res[1] = i;flag = false;}else{res[1] = i;}}}return res;}
}

四、力扣922. 按奇偶排序数组 II

class Solution {public int[] sortArrayByParityII(int[] nums) {int len = nums.length;int[] arr1 = new int[len];int[] arr2 = new int[len];for(int i = 0, j = 0, k = 1; i < len; i ++){if(nums[i]%2 == 0){arr1[j] = nums[i];j += 2;}else{arr2[k] = nums[i];k += 2;}}for(int i = 0; i < len ; i++){if(i % 2 == 0){nums[i] = arr1[i];}else{nums[i] = arr2[i];}}return nums;}
}