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sortColors

时间:2019-09-16 00:22:37      阅读:110      评论:0      收藏:0      [点我收藏+]

标签:length   需要   object   span   red   链接   商业   fun   处理   

Given an array with n objects colored red, white or blue, sort them in-place so that objects of the same color are adjacent, with the colors in the order red, white and blue.

Here, we will use the integers 0, 1, and 2 to represent the color red, white, and blue respectively.

Note: You are not suppose to use the library‘s sort function for this problem.

Example:

Input: [2,0,2,1,1,0]
Output: [0,0,1,1,2,2]

来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/sort-colors
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。

 

解法1:遍历元素,把0调到数组的前面,把2调到数组后面,1不用处理:

 public void sortColors(int[] nums) {
        int length = nums.length;
        int temp,index;
        index=0;
        for(int i = 0; i < length; i++)
        {
            if(nums[i] == 0)
            {
                temp = nums[index];
                nums[index++] = nums[i];
                nums[i] = temp;
            }
        }
        index=length - 1;
//这里其实可以加个限定条件,因为是从数组末端开始遍历的,当遍历到0的时候,就已经不用再往前遍历了,因此也可节省一丁点时间,不过好像问题并不大,需要优化的是内存方面。。。
for(int j = length - 1; j >= 0; j--) { if(nums[j] == 2) { temp = nums[index]; nums[index--] = nums[j]; nums[j] = temp; } } }

解法2:遍历计数0,1,2的个数,再把原数组重新赋值:

    public void sortColors(int[] nums) {
        int length = nums.length;
        int count = 0;
        int count1 = 0;
        int count2;
        for(int i = 0; i < length; i++)
        {
            if(nums[i] == 0)
            {
                nums[count] = 0;
                count++;
            }
            else if(nums[i] == 1)
            {
                count1++;
            }
        }
        count2 = length - (count + count1);
        
        for(int i = count; i < count + count1; i++)
        {
            nums[i] = 1;
        }
        for(int i = count+count1; i < length; i++)
        {
            nums[i] = 2;
        }
    }

 

sortColors

标签:length   需要   object   span   red   链接   商业   fun   处理   

原文地址:https://www.cnblogs.com/WakingShaw/p/11524992.html

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