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Merge pull request #1470 from wzqwtt/greedy04
添加(1005.K次取反后最大化的数组和、0134.加油站) Scala版本
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@ -471,5 +471,73 @@ int canCompleteCircuit(int* gas, int gasSize, int* cost, int costSize){
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}
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```
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### Scala
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暴力解法:
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```scala
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object Solution {
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def canCompleteCircuit(gas: Array[Int], cost: Array[Int]): Int = {
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for (i <- cost.indices) {
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var rest = gas(i) - cost(i)
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var index = (i + 1) % cost.length // index为i的下一个节点
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while (rest > 0 && i != index) {
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rest += (gas(index) - cost(index))
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index = (index + 1) % cost.length
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}
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if (rest >= 0 && index == i) return i
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}
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-1
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}
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}
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```
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贪心算法,方法一:
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```scala
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object Solution {
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def canCompleteCircuit(gas: Array[Int], cost: Array[Int]): Int = {
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var curSum = 0
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var min = Int.MaxValue
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for (i <- gas.indices) {
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var rest = gas(i) - cost(i)
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curSum += rest
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min = math.min(min, curSum)
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}
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if (curSum < 0) return -1 // 情况1: gas的总和小于cost的总和,不可能到达终点
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if (min >= 0) return 0 // 情况2: 最小值>=0,从0号出发可以直接到达
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// 情况3: min为负值,从后向前看,能把min填平的节点就是出发节点
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for (i <- gas.length - 1 to 0 by -1) {
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var rest = gas(i) - cost(i)
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min += rest
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if (min >= 0) return i
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}
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-1
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}
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}
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```
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贪心算法,方法二:
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```scala
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object Solution {
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def canCompleteCircuit(gas: Array[Int], cost: Array[Int]): Int = {
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var curSum = 0
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var totalSum = 0
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var start = 0
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for (i <- gas.indices) {
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curSum += (gas(i) - cost(i))
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totalSum += (gas(i) - cost(i))
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if (curSum < 0) {
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start = i + 1 // 起始位置更新
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curSum = 0 // curSum从0开始
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}
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}
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if (totalSum < 0) return -1 // 说明怎么走不可能跑一圈
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start
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}
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}
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```
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-----------------------
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<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>
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@ -289,6 +289,28 @@ function largestSumAfterKNegations(nums: number[], k: number): number {
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};
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```
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### Scala
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```scala
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object Solution {
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def largestSumAfterKNegations(nums: Array[Int], k: Int): Int = {
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var num = nums.sortWith(math.abs(_) > math.abs(_))
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var kk = k // 因为k是不可变量,所以要赋值给一个可变量
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for (i <- num.indices) {
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if (num(i) < 0 && kk > 0) {
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num(i) *= -1 // 取反
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kk -= 1
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}
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}
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// kk对2取余,结果为0则为偶数不需要取反,结果为1为奇数,只需要对最后的数字进行反转就可以
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if (kk % 2 == 1) num(num.size - 1) *= -1
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num.sum // 最后返回数字的和
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}
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}
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```
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