Candy
There are N children standing in a line. Each child is assigned a rating value.
You are giving candies to these children subjected to the following requirements:
- Each child must have at least one candy.
- Children with a higher rating get more candies than their neighbors.
What is the minimum candies you must give?
贪心,想通了两轮遍历完事。
第一轮第一个小盆与先给1根,从左往右扫,如果发现当前的rating小于左边的rating,给的糖比左边的多1。
第二轮从倒数第二个小盆与开始,从右往左扫,如果发现当前的rating小于右边的rating,并且糖也少,给的糖比左边的多1。
1 /** 2 * @param {number[]} ratings 3 * @return {number} 4 */ 5 var candy = function(ratings) { 6 var i, result; rates = []; 7 rates[0] = 1; 8 for(i = 1; i < ratings.length; i++){ 9 if(ratings[i] > ratings[i - 1]){10 rates[i] = rates[i - 1] + 1;11 }else{12 rates[i] = 1;13 }14 }15 result = rates[ratings.length - 1];16 for(i = ratings.length - 2; i >= 0; i--){17 if(ratings[i] > ratings[i + 1] && rates[i] <= rates[i + 1]){18 rates[i] = rates[i + 1] + 1;19 }20 result += rates[i];21 }22 return result;23 };
一开始想了一个解法,有点费劲。
第一轮遍历,如果这个小盆与比左右的rating都要小,他肯定是1,把index加入到queue中。
然后就开始递归,有点像bfs。
这样递归的话,保证大的rating一定后做,先处理的是较小的rating。
递归的时候分两种情况:
1. 当前ratings小于左边且大于右边 或者 当前ratings小于右边且大于左边。
因为大的一定后做,所以可以确定当前点的糖数,等于max(左边糖,右边糖) + 1,注意这里如果还没有给过糖,默认值是0,较大的一边是0。
2. 如果两边都给过糖了,并且当前ratings大于左边且大于右边,当前节点的糖数也是等于max(左边糖,右边糖) + 1。
这一轮给完糖后,把左边右边和当前的没有发过糖的节点都塞进queue中。
虽然比较复杂,时间复杂度仍然是O(n)。
但是交上去之后12000的那组数据堆栈溢出。
唉,本地测试10几毫秒妥妥的,leetcode的JS小堆栈居然爆了,是在下输了。
1 /** 2 * @param {number[]} ratings 3 * @return {number} 4 */ 5 var candyMLE = function(ratings) { 6 var queue = [], result = 0, rates = []; 7 for(var i = 0; i < ratings.length; i++){ 8 if(ratings[i - 1] && ratings[i - 1] < ratings[i]){ 9 continue;10 }11 if(ratings[i + 1] && ratings[i + 1] < ratings[i]){12 continue;13 }14 queue.push(i);15 result++;16 rates[i] = 1;17 }18 bfs(queue);19 return result;20 21 function bfs(queue){22 var len = queue.length;23 if(queue.length !== 0){24 while(len--){25 var top = queue.shift();26 if(!rates[top]){27 giveCandy(queue, top);28 }29 if(!rates[top - 1] && ratings[top - 1] && queue.indexOf(top - 1) === -1){30 queue.push(top - 1);31 }32 if(!rates[top + 1] && ratings[top + 1] && queue.indexOf(top + 1) === -1){33 queue.push(top + 1);34 } 35 if(!rates[top] && queue.indexOf(top) === -1){36 queue.push(top);37 } 38 }39 bfs(queue); 40 }41 }42 function giveCandy(queue, i){43 if(ratings[i]){ 44 var leftValue = ratings[i - 1] || 0;45 var leftRate = rates[i - 1] || 0;46 var rightValue = ratings[i + 1] || 0;47 var rightRate = rates[i + 1] || 0;48 if(ratings[i] > leftValue && ratings[i] < rightValue || 49 ratings[i] < leftValue && ratings[i] > rightValue ||50 ratings[i] >= leftValue && ratings[i] >= rightValue && (leftRate !== 0 || !ratings[i - 1])){51 rates[i] = Math.max(leftRate, rightRate) + 1;52 result += rates[i];53 }54 }55 }56 };