代码随想录三刷day55

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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;
    }
}