leetcode: Summary Ranges
2016-01-06 13:02
232 查看
思路:为方便运算,将给长度为len的给定数组nums扩展为长度为len+1的tmpNums,最后一个元素只要比其前面的元素差值超过1即可。记录起始m,终止i。先判断是否连续。不连续:再比较m与i-1是否是同一位置如果是,则直接加入输出列表;如果不是,则将i-1位置上的元素加入字符串,并加入到输出列表。连续:直接扫描下一个元素
代码:
public class Solution {
public List<String> summaryRanges(int[] nums) {
List<String> res = new ArrayList<String>();
int len = nums.length;
int i=1;
int[] tmpNums = new int[len+1];
for(int j=0; j<len; j++){
tmpNums[j] = nums[j];
}
if(len == 0) {
return res;
}
tmpNums[len] = nums[len-1]+3;
if(len == 1) {
res.add(tmpNums[0]+"");
return res;
}
StringBuilder sb = new StringBuilder(tmpNums[i-1]+"");
int m = i-1;
while(i<len+1) {
if(tmpNums[i] != tmpNums[i-1] + 1) {
if(m == i-1) {
res.add(sb.toString());
} else {
res.add(sb.append("->"+tmpNums[i-1]).toString());
}
sb = new StringBuilder(tmpNums[i]+"");
m = i;
}
i++;
}
return res;
}
}
代码:
public class Solution {
public List<String> summaryRanges(int[] nums) {
List<String> res = new ArrayList<String>();
int len = nums.length;
int i=1;
int[] tmpNums = new int[len+1];
for(int j=0; j<len; j++){
tmpNums[j] = nums[j];
}
if(len == 0) {
return res;
}
tmpNums[len] = nums[len-1]+3;
if(len == 1) {
res.add(tmpNums[0]+"");
return res;
}
StringBuilder sb = new StringBuilder(tmpNums[i-1]+"");
int m = i-1;
while(i<len+1) {
if(tmpNums[i] != tmpNums[i-1] + 1) {
if(m == i-1) {
res.add(sb.toString());
} else {
res.add(sb.append("->"+tmpNums[i-1]).toString());
}
sb = new StringBuilder(tmpNums[i]+"");
m = i;
}
i++;
}
return res;
}
}
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