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; } }
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
