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res.js
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90 lines (83 loc) · 2.21 KB
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/**
* Given an array with n objects colored red, white or blue, sort them so that objects of the same color are adjacent, with
*
* 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.
*
* Follow up:
* A rather straight forward solution is a two-pass algorithm using counting sort.
*
* First, iterate the array counting number of 0's, 1's, and 2's, then overwrite array with total number of 0's, then 1's and followed by 2's.
*
* Could you come up with an one-pass algorithm using only constant space?
*
* res.js
* @authors Joe Jiang ([email protected])
* @date 2017-03-02 23:33:12
* @version $Id$
*
* @param {number[]} nums
* @return {void} Do not return anything, modify nums in-place instead.
*/
let sortColors = function(nums) {
let numsize = nums.length,
begin = 0,
end = numsize-1;
for (let i=0; i<=end; i++) {
let currentVal = nums[i];
switch (currentVal) {
case 0:
nums[i] = nums[begin];
nums[begin++] = 0;
break;
case 2:
nums[i--] = nums[end];
nums[end--] = 2;
break;
default:
break;
}
}
};
// another one-pass solution
let sortColors = function(nums) {
let numsize = nums.length,
n0 = -1,
n1 = -1,
n2 = -1;
for (let i = 0; i < numsize; ++i) {
if (nums[i] == 0) {
nums[++n2] = 2;
nums[++n1] = 1;
nums[++n0] = 0;
} else if (nums[i] == 1) {
nums[++n2] = 2;
nums[++n1] = 1;
} else if (nums[i] == 2) {
nums[++n2] = 2;
}
}
};
// traditional two-pass solution
let sortColors = function(nums) {
let numsize = nums.length,
n0 = 0,
n1 = 0;
for (let i = 0; i < numsize; i++) {
if (nums[i] === 0) {
nums[n0++] = 0;
} else if (nums[i] === 1) {
n1 += 1;
}
}
n1 += n0;
for (let i = n0; i < numsize; i++) {
if (i<n1) {
nums[i] = 1;
} else {
nums[i] = 2;
}
}
};